Counting and Probability Practice Questions — GMAT

Question 1Question →

A bag contains 5 red marbles and 5 blue marbles. If 2 marbles are randomly selected from the bag one after another without replacement, what is the probability that at least one of the selected marbles is red?

\frac{7}{9}

\frac{2}{9}

\frac{5}{9}

\frac{3}{4}

\frac{1}{4}

Show answer & explanation

Answer: $\frac{7}{9}$

Answer

\frac{7}{9}

To find the probability of drawing at least one red marble, calculate the probability of the complementary event (drawing two blue marbles) and subtract it from 1. The probability of selecting a blue marble on the first draw is

\frac{5}{10}

, and on the second draw without replacement it is

\frac{4}{9}

. The probability of both marbles being blue is

\frac{5}{10} \times \frac{4}{9} = \frac{2}{9}

. Subtracting this from 1 gives

1 - \frac{2}{9} = \frac{7}{9}

.

Step-by-Step Solution

1

Identify the complementary outcome

The event 'at least one marble is red' is complementary to the event 'no marbles are red' (i.e., 'both marbles are blue').

Using the relationship

P(\text{at least one red}) = 1 - P(\text{both blue})

is simpler than summing individual favorable cases.

2

Calculate the probability of drawing two blue marbles without replacement

The probability that the first marble is blue is

\frac{5}{10} = \frac{1}{2}

. The probability that the second marble is blue is

\frac{4}{9}

. Therefore,

P(\text{both blue}) = \frac{5}{10} \times \frac{4}{9} = \frac{20}{90} = \frac{2}{9}

.

Since draws are without replacement, the sample space and number of blue marbles decrease by 1 after the first draw.

3

Subtract the complementary probability from 1

P(\text{at least one red}) = 1 - \frac{2}{9} = \frac{7}{9}

.

The sum of complementary probabilities is always equal to 1.

Key Concept

Complementary Probability:

P(\text{at least one}) = 1 - P(\text{none})

Question 2Question →

A wooden box contains 15 identical tokens, each marked with a distinct integer from $1$ to $15$ , inclusive. If one token is drawn at random from the box, what is the probability that the integer on the drawn token is a prime number?

\frac{1}{3}

\frac{2}{5}

\frac{7}{15}

\frac{8}{15}

\frac{2}{3}

Show answer & explanation

Answer: $\frac{2}{5}$

Answer

The probability that the integer on the drawn token is a prime number is

\frac{2}{5}

.

The total number of possible outcomes when choosing one token from

15

tokens is

15

. The prime numbers between

1

and

15

, inclusive, are

2, 3, 5, 7, 11,

and

13

. There are

6

favorable outcomes. The probability is therefore

\frac{6}{15}

, which simplifies to

\frac{2}{5}

.

Step-by-Step Solution

1

Determine the total number of possible outcomes (the denominator).

The total number of tokens is

15

, so

N = 15

.

The sample space consists of all integers from

1

to

15

, inclusive.

2

Identify and count all prime numbers in the set from

1

to

15

, inclusive.

The prime numbers in this set are

2, 3, 5, 7, 11,

and

13

. Thus, there are

6

prime numbers.

A prime number is an integer strictly greater than

1

that has exactly two distinct positive divisors:

1

and itself. Note that

1

is not prime.

3

Calculate the basic probability

P(\text{Prime}) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}

.

P(\text{Prime}) = \frac{6}{15} = \frac{2}{5}

.

Dividing the favorable count

6

by total count

15

and simplifying by dividing numerator and denominator by

3

yields

\frac{2}{5}

.

Key Concept

Basic Single-Event Probability and Definition of Prime Numbers

Estimated Time:45s

Question 3Question →

An analyst at an investment firm evaluates a portfolio of $160$ corporate bonds for two potential risk factors: credit rating downgrade risk and liquidity risk. The evaluation reveals that $60$ bonds have credit rating downgrade risk, $72$ bonds have liquidity risk, and $64$ bonds have neither risk factor. If a bond is selected at random from those in the portfolio that have at least one of the two risk factors, what is the probability that it has credit rating downgrade risk? Express your answer as a decimal.

Show answer & explanation

Answer: 0.625

Answer

The probability is 0.625 (or 5/8).

To calculate the probability that a bond has credit rating downgrade risk given that it has at least one risk factor, the sample space must be restricted to bonds with at least one risk factor. Out of

160

bonds,

64

have neither risk factor, leaving

160 - 64 = 96

bonds with at least one risk factor. All

60

bonds with credit rating downgrade risk are part of this group. The required conditional probability is

\frac{60}{96} = \frac{5}{8} = 0.625

.

Step-by-Step Solution

1

Calculate the size of the restricted sample space (bonds with at least one risk factor).

Total bonds with at least one risk factor = 160 - 64 = 96 bonds.

The condition specifies that the selection is made only from bonds having at least one risk factor.

2

Identify the number of favorable outcomes within this restricted sample space.

Number of bonds with credit rating downgrade risk = 60.

All 60 bonds with credit rating downgrade risk inherently possess at least one risk factor, so they lie entirely within the restricted sample space.

3

Compute the conditional probability P(Downgrade Risk | At Least One Risk).

60 / 96 = 5 / 8 = 0.625.

Conditional probability requires dividing the count of favorable outcomes by the count of the restricted sample space.

Key Concept

Conditional Probability and Sample Space Restriction

Question 4Question →

A clinical study evaluated $150$ patient records for the presence of three specific health biomarkers: Biomarker X, Biomarker Y, and Biomarker Z. The study revealed that $70$ patients had Biomarker X, $60$ patients had Biomarker Y, and $50$ patients had Biomarker Z. Exactly $35$ patients had exactly two of these biomarkers, and exactly $10$ patients had all three biomarkers. How many of the patients evaluated had none of the three biomarkers?

5

15

25

35

45

Show answer & explanation

Answer: 25

Answer

25 patients evaluated had none of the three biomarkers.

By the inclusion-exclusion principle for three sets, the total number of elements in at least one set is given by

|X \cup Y \cup Z| = |X| + |Y| + |Z| - (\text{Elements in exactly two sets}) - 2(\text{Elements in all three sets})

. Substituting the given values yields

70 + 60 + 50 - 35 - 2(10) = 180 - 35 - 20 = 125

. Since there are

150

total patients evaluated, the number of patients with none of the three biomarkers is

150 - 125 = 25

.

Step-by-Step Solution

1

Sum the individual counts for all three biomarker groups.

\text{Sum of individual sets} = 70 + 60 + 50 = 180

This sum counts individuals with 1 biomarker once, individuals with 2 biomarkers twice, and individuals with 3 biomarkers three times.

2

Apply the inclusion-exclusion formula adapted for 'exactly two' and 'all three' overlapping subsets.

\text{At least one biomarker} = 180 - (\text{Exactly two}) - 2 \times (\text{All three}) = 180 - 35 - 2(10) = 125

To count each person with at least one biomarker exactly once, we subtract the count of patients with exactly two biomarkers once, and the count of patients with all three biomarkers twice.

3

Subtract the number of patients with at least one biomarker from the total population of patients evaluated.

\text{None} = 150 - 125 = 25

The total group consists of patients with at least one biomarker plus patients with none of the three biomarkers.

Key Concept

Three-Set Overlapping Sets and Inclusion-Exclusion Principle

Estimated Time:1m 45s

Question 5Question →

Four distinct books—two mathematics books and two history books—are to be arranged in a single line on a shelf. If the two mathematics books must stand next to each other, in how many different linear arrangements can the four books be placed?

Show answer & explanation

Answer: 12

Answer

12 linear arrangements

To find the total number of linear arrangements where the two mathematics books are adjacent, treat the two mathematics books as a single combined block. This leaves 3 items to arrange (the mathematics block and the 2 individual history books), which can be arranged in

3! = 6

ways. Within the block, the 2 mathematics books can be ordered in

2! = 2

ways. Multiplying the independent arrangements yields

6 \times 2 = 12

total arrangements.

Step-by-Step Solution

1

Group the adjacent items into a single unit and count total units to arrange

3 units (1 grouped math unit + 2 individual history books), which can be arranged in

3! = 6

ways

Treating items that must be adjacent as a single block simplifies the linear arrangement into distinct available positions.

2

Determine internal permutations within the grouped unit

2! = 2

ways to order the two mathematics books inside their block

The mathematics books can switch positions with each other within their designated block.

3

Multiply external and internal arrangements

6 \times 2 = 12

total linear arrangements

By the Fundamental Counting Principle, total arrangements equal the product of unit arrangements and internal arrangements.

Key Concept

Linear Permutations with Adjacent Restrictions (Tie-Together Method)

Question 6Question →

An interior designer is arranging a row of 7 decorative wall tiles consisting of 3 identical blue tiles, 2 identical yellow tiles, and 2 identical red tiles. If the 2 red tiles cannot be placed next to each other, how many distinct arrangements of the 7 tiles are possible?

Show answer & explanation

Answer: 150

Answer

The total number of distinct arrangements possible is 150.

To find the total number of distinct arrangements where no two red tiles are adjacent, first calculate the arrangements of the 5 non-restricted tiles (3 blue, 2 yellow), which is

\frac{5!}{3!2!} = 10

. Placing 5 tiles creates 6 available spaces (including the two ends). Selecting 2 of these 6 spaces for the 2 identical red tiles yields

\binom{6}{2} = 15

choices. By the Fundamental Counting Principle, the total number of valid arrangements is

10 \times 15 = 150

.

Step-by-Step Solution

1

Calculate the number of distinct ways to arrange the non-restricted tiles (3 identical blue and 2 identical yellow).

The number of distinct arrangements of the 5 non-red tiles is

\frac{5!}{3!2!} = \frac{120}{6 \times 2} = 10

.

Arranging all non-restricted tiles first creates the specific positions into which the restricted tiles can be inserted.

2

Determine the number of valid positions for the 2 identical red tiles such that no two are adjacent.

Placing 5 tiles creates 6 available insertion slots (one at each end and four between tiles). Choosing 2 distinct slots out of 6 gives

\binom{6}{2} = \frac{6 \times 5}{2} = 15

ways.

Selecting 2 distinct slots ensures that every selected space holds at most one red tile, guaranteeing that no two red tiles are adjacent.

3

Multiply the results from Step 1 and Step 2 using the Fundamental Counting Principle.

10 \times 15 = 150

.

Each arrangement of non-red tiles can be independently paired with any valid placement of the red tiles.

Key Concept

Counting arrangements of identical items with non-adjacency restrictions using the slotting method.

Estimated Time:1m 30s

Question 7Question →

How many distinct 4-letter arrangements can be formed by rearranging all of the letters in the word $\text{SEES}$ ?

Show answer & explanation

Answer: 6

Answer

6

To calculate the number of unique arrangements for a multiset of letters, compute the total factorial divided by the product of factorials for each repeated letter's frequency. For

\text{SEES}

, there are 4 total letters with 2 'S's and 2 'E's, resulting in

\frac{4!}{2!2!} = \frac{24}{4} = 6

.

Step-by-Step Solution

1

Count the total number of letters and identify frequencies of repeated letters.

The word

\text{SEES}

contains 4 letters in total: two 'S's and two 'E's.

To apply the distinct permutations formula, we need the total count of elements and the counts for each repeated identical element.

2

Calculate the number of distinct arrangements using the formula

\frac{n!}{n_1! n_2! \dots n_k!}

.

\frac{4!}{2! \cdot 2!} = \frac{24}{2 \cdot 2} = 6.

Dividing by

2! \cdot 2!

eliminates duplicate counts arising from swapping indistinguishable identical letters.

Key Concept

Permutations with Repetition

Question 8Question →

A project committee of $4$ members is to be selected from a pool of $n$ senior engineers and $6$ junior engineers, where $n \ge 4$ . The committee must contain at least one senior engineer and at least one junior engineer. If there are exactly $310$ different possible ways to select the committee, what is the value of $n$ ?

4

5

6

7

8

Show answer & explanation

Answer: 5

Answer

The value of

n

is 5.

Using complementary counting, the total number of ways to pick any 4 members from the

(n+6)

total engineers is

\binom{n+6}{4}

. The condition requires at least one senior engineer and at least one junior engineer. The invalid cases are when all 4 are junior engineers (which can occur in

\binom{6}{4} = 15

ways) or all 4 are senior engineers (which can occur in

\binom{n}{4}

ways). Setting up the equation:

\binom{n+6}{4} - \binom{6}{4} - \binom{n}{4} = 310

, which simplifies to

\binom{n+6}{4} - \binom{n}{4} = 325

. Testing

n = 5

gives

\binom{11}{4} - \binom{5}{4} = 330 - 5 = 325

, which matches the given condition.

Step-by-Step Solution

1

Set up the combination formula for total unrestricted selections

Total selections from

(n + 6)

engineers choosing 4 is

\binom{n+6}{4}

.

Order of selection does not matter when forming a committee.

2

Identify the restricted (invalid) committee configurations

Committees with 0 senior engineers:

\binom{6}{4} = 15

. Committees with 0 junior engineers:

\binom{n}{4}

.

The committee must contain at least one member from each category.

3

Formulate the equation using complementary counting

\binom{n+6}{4} - \binom{n}{4} - 15 = 310 \implies \binom{n+6}{4} - \binom{n}{4} = 325

.

Subtracting invalid configurations from total configurations yields the valid configurations.

4

Test values for

n

For

n = 5

:

\binom{11}{4} - \binom{5}{4} = 330 - 5 = 325

.

Since

325 = 325

,

n = 5

satisfies the given constraint exactly.

Key Concept

Group selections with constraints using complementary counting:

\text{Valid} = \text{Total} - \text{Restricted}

.

Question 9Question →

A quality assurance technician randomly selects 2 microprocessors from a batch of $16$ microprocessors, of which $6$ were manufactured at Facility A and $10$ were manufactured at Facility B. The selections are made one after another without replacement. What is the probability that both selected microprocessors were manufactured at Facility A?

Show answer & explanation

Answer: 0.125

Answer

The probability that both selected microprocessors were manufactured at Facility A is 0.125 (or 1/8).

Because the selections are made without replacement, the outcome of the first draw changes the pool of available microprocessors for the second draw. The probability of choosing a Facility A microprocessor first is

\frac{6}{16} = \frac{3}{8}

. Following that selection,

5

Facility A microprocessors remain among

15

total microprocessors, giving a conditional probability of

\frac{5}{15} = \frac{1}{3}

for the second selection. Multiplying these dependent probabilities yields

\frac{3}{8} \times \frac{1}{3} = \frac{1}{8} = 0.125

.

Step-by-Step Solution

1

Determine the probability of selecting a Facility A microprocessor on the first draw.

P(\text{First is Facility A}) = \frac{6}{16} = \frac{3}{8}

There are 6 microprocessors from Facility A out of 16 total microprocessors.

2

Determine the conditional probability of selecting a Facility A microprocessor on the second draw.

P(\text{Second is Facility A} \mid \text{First is Facility A}) = \frac{5}{15} = \frac{1}{3}

Since the selection is made without replacement, 5 Facility A microprocessors remain out of a reduced total of 15 microprocessors.

3

Multiply the dependent probabilities to find the combined probability.

P(\text{Both are Facility A}) = \frac{3}{8} \times \frac{1}{3} = \frac{3}{24} = \frac{1}{8} = 0.125

The joint probability of sequential dependent events is the product of the initial probability and the conditional probability.

Key Concept

Probability of Dependent Events (Sampling without Replacement)

Estimated Time:1m 30s

Question 10Question →

A panel of 6 distinct experts—3 scientists, 2 economists, and 1 moderator—are to sit in a single row of 6 chairs for a discussion. If the 3 scientists must all sit in adjacent chairs and the 2 economists cannot sit in adjacent chairs, how many different seating arrangements are possible?

12

36

72

108

144

Show answer & explanation

Answer: 72

Answer

72 seating arrangements

To find the number of valid seating arrangements, we first group the 3 scientists together as 1 unit, leaving us with 2 non-economist entities (the scientist block and the moderator). These 2 entities can be arranged in 2! = 2 ways. Placing the 2 distinct economists into the 3 available gaps around these entities ensures they are not adjacent, yielding P(3, 2) = 6 choices. Finally, multiplying by the 3! = 6 internal arrangements of the scientists yields a total of 2 * 6 * 6 = 72 valid arrangements.

Step-by-Step Solution

1

Group the 3 scientists into a single block

S

, and treat the moderator

M

as an individual unit.

There are 2 non-economist units: block

S

and moderator

M

.

Grouping elements that must be adjacent allows us to treat them temporarily as a single entity.

2

Calculate the arrangements of the non-economist units.

The 2 non-economist units can be arranged in

2! = 2

ways.

Linear arrangement of 2 distinct entities.

3

Insert the 2 economists into the available gaps created by the non-economist units.

For any arrangement of

S

and

M

(e.g., _

S

_

M

_), there are 3 available gaps. The 2 distinct economists can be placed in these gaps in

P(3,2) = 3 \times 2 = 6

ways.

To ensure no two economists sit together, each economist must occupy a separate gap.

4

Account for the internal arrangements of the 3 scientists within block

S

.

The 3 distinct scientists can be arranged among themselves in

3! = 6

ways.

Order matters among distinct individuals within a grouped block.

5

Apply the fundamental counting principle to compute total arrangements.

2 \times 6 \times 6 = 72

total arrangements.

Multiply the independent choices made in steps 2, 3, and 4.

Key Concept

Permutations with Adjacency and Non-Adjacency Restrictions

Question 11Question →

An integer is selected at random from the set of all positive integers less than or equal to 40. Given that the selected integer is a multiple of 3, what is the probability that it is also a multiple of 4?

\frac{3}{40}

\frac{3}{13}

\frac{3}{10}

\frac{1}{4}

\frac{4}{13}

Show answer & explanation

Answer: $\frac{3}{13}$

Answer

The probability that the selected integer is a multiple of 4, given that it is a multiple of 3, is

\frac{3}{13}

.

The condition 'given that the selected integer is a multiple of 3' restricts the sample space to the 13 positive integers up to 40 that are divisible by 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, and 39. Among these 13 numbers, those that are also multiples of 4 are multiples of 12, namely 12, 24, and 36 (3 numbers). Therefore, the conditional probability is

\frac{3}{13}

.

Step-by-Step Solution

1

Identify the restricted sample space defined by the condition.

The positive integers less than or equal to 40 that are multiples of 3 are: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, and 39. There are 13 such integers.

For conditional probability

P(A \mid B)

, the sample space must be restricted to all outcomes satisfying condition

B

(being a multiple of 3).

2

Identify the favorable outcomes within the restricted sample space.

Integers that are multiples of both 3 and 4 must be multiples of 12. Among the 13 multiples of 3, those that are also multiples of 12 are: 12, 24, and 36. There are 3 such integers.

The numerator of

P(A \mid B)

counts the elements in the intersection

A \cap B

.

3

Calculate the conditional probability.

P(\text{Multiple of 4} \mid \text{Multiple of 3}) = \frac{3}{13}

.

Divide the number of favorable outcomes (3) by the total number of outcomes in the restricted sample space (13).

Key Concept

Conditional Probability with Restricted Sample Space

Estimated Time:1m 30s

Question 12Question →

A laboratory tested $200$ synthetic compound samples for two properties: thermal stability and chemical resistance. Among the samples, $120$ exhibited thermal stability, $90$ exhibited chemical resistance, and $50$ exhibited neither property. If a sample is selected at random from those that exhibited thermal stability, what is the probability that it also exhibited chemical resistance?

Show answer & explanation

Answer: 0.5

Answer

0.5

The conditional probability of selecting a sample with chemical resistance given that it has thermal stability is found by dividing the number of samples with both properties (

60

) by the total number of samples with thermal stability (

120

), giving

\frac{60}{120} = 0.5

.

Step-by-Step Solution

1

Find the total number of samples exhibiting at least one of the two properties.

Since

50

out of

200

samples exhibited neither property, the number of samples exhibiting at least one property is

200 - 50 = 150

.

The total population consists of samples exhibiting at least one property plus samples exhibiting neither property.

2

Calculate the number of samples exhibiting both thermal stability (

T

) and chemical resistance (

C

).

Using the inclusion-exclusion principle

|T \cup C| = |T| + |C| - |T \cap C|

, we have

150 = 120 + 90 - |T \cap C|

, which yields

|T \cap C| = 60

.

Overlapping sets require subtracting the intersection to avoid double-counting elements.

3

Compute the conditional probability

P(C|T)

.

P(C|T) = \frac{|T \cap C|}{|T|} = \frac{60}{120} = 0.5

.

The given condition restricts the sample space to only the

120

samples exhibiting thermal stability.

Key Concept

Conditional probability restricts the sample space to the given condition's outcome space:

P(A|B) = \frac{|A \cap B|}{|B|}

.

Estimated Time:1m 30s

Question 13Question →

A logistics firm generates a 4-character tracking code, $c_1 c_2 c_3 c_4$ , for express parcels according to the following rules:
- The first character, $c_1$ , must be a non-zero digit ( $1$ through $9$ ).
- The second character, $c_2$ , must be a vowel chosen from the set $\{\text{A, E, I, O, U}\}$ .
- The third character, $c_3$ , must be an even digit ( $0, 2, 4, 6, 8$ ).
- The fourth character, $c_4$ , must be a consonant chosen from the set $\{\text{B, F, G, K, M, P}\}$ .

If no digit may be repeated within the code, how many different tracking codes can be created?

Show answer & explanation

Answer: 1230

Answer

1230

To find the total number of unique tracking codes, calculate the independent choices for letters and dependent choices for digits. For digits

c_1

and

c_3

, if

c_1

is odd (

1, 3, 5, 7, 9

), there are

5 \times 5 = 25

ways to choose

c_3

. If

c_1

is even (

2, 4, 6, 8

), there are

4 \times 4 = 16

ways to choose

c_3

, giving

41

valid digit pairs. Multiplying by the

5

vowel options for

c_2

and

6

consonant options for

c_4

yields

41 \times 5 \times 6 = 1230

.

Step-by-Step Solution

1

Determine valid digit choices for

c_1

and

c_3

considering the non-repetition constraint

Split into Case 1 (

c_1

is odd) and Case 2 (

c_1

is even)

The availability of even digits for

c_3

depends on whether

c_1

uses an even or odd digit.

2

Calculate the number of ways to pick

c_1

and

c_3

Case 1 (

c_1 \in \{1, 3, 5, 7, 9\}

):

5 \times 5 = 25

ways. Case 2 (

c_1 \in \{2, 4, 6, 8\}

):

4 \times 4 = 16

ways. Total digit combinations =

25 + 16 = 41

.

If

c_1

is odd, all 5 even digits remain for

c_3

. If

c_1

is even, only 4 even digits remain for

c_3

.

3

Calculate the number of ways to pick letters

c_2

and

c_4

5 choices for

c_2

and 6 choices for

c_4

, giving

5 \times 6 = 30

letter combinations.

Vowels and consonants are chosen independently from distinct sets without restrictions between them.

4

Multiply independent stage outcomes using the Fundamental Counting Principle

41 \times 30 = 1230

The overall number of distinct codes is the product of digit pair combinations and letter pair combinations.

Key Concept

Fundamental Counting Principle with Conditional Restrictions

Question 14Question →

A chef is arranging 6 distinct dishes—3 appetizers and 3 main courses—in a single straight row on a serving table for a culinary presentation. If all 3 appetizers must be placed adjacent to one another and all 3 main courses must also be placed adjacent to one another, in how many different linear arrangements can the dishes be displayed?

18

36

72

144

720

Show answer & explanation

Answer: 72

Answer

72

To find the number of valid linear arrangements, treat the 3 appetizers as a single group and the 3 main courses as another single group. There are

2! = 2

ways to arrange these two groups on the table (appetizers first or main courses first). Within the appetizer group, the 3 distinct dishes can be arranged in

3! = 6

ways. Within the main course group, the 3 distinct dishes can also be arranged in

3! = 6

ways. By the Fundamental Counting Principle, the total number of linear arrangements is

2! \times 3! \times 3! = 2 \times 6 \times 6 = 72

.

Step-by-Step Solution

1

Group the dishes into distinct blocks based on the given adjacency restriction.

Create Block A (containing all 3 appetizers) and Block M (containing all 3 main courses).

Because items of the same category must remain adjacent, each category functions as a single compound element.

2

Calculate the number of ways to arrange the 2 blocks relative to each other.

Number of block arrangements =

2! = 2

.

The two blocks can be placed in order (Block A then Block M) or (Block M then Block A).

3

Calculate the internal permutations within each block.

Internal arrangements for appetizers =

3! = 6

, and internal arrangements for main courses =

3! = 6

.

Each category consists of 3 distinct dishes that can be permuted among themselves.

4

Apply the Fundamental Counting Principle to find total linear arrangements.

Total arrangements =

2! \times 3! \times 3! = 2 \times 6 \times 6 = 72

.

The decision steps are independent, so their possible outcomes are multiplied.

Key Concept

Permutations with Block Restrictions (Grouping Method)

Practice More

Try solving a problem where only one specific subset of items must remain together while the rest can be arranged freely.

Alternative Method

Alternatively, place the first appetizer in slot 1 (3 choices). The remaining appetizers must take slots 2 and 3 (2 choices and 1 choice). Then slots 4, 5, 6 must be filled by main courses (3 choices, 2 choices, 1 choice). This gives

3 \times 2 \times 1 \times 3 \times 2 \times 1 = 36

ways when appetizers are placed first. Symmetrically, placing main courses in slots 1 to 3 gives another 36 ways, totaling

36 + 36 = 72

ways.

Estimated Time:1m 30s

Question 15Question →

A task force must select a 4-member subcommittee from a pool consisting of 6 architects and $n$ urban planners. If there are exactly 111 different possible subcommittees that include at least 1 urban planner, what is the value of $n$ ?

2

3

4

5

9

Show answer & explanation

Answer: 3

Answer

The value of

n

is 3.

Using the complementary counting method, the number of subcommittees with at least 1 urban planner equals the total possible 4-member subcommittees from the pool of

(6+n)

people minus the subcommittees formed entirely of architects. The number of all-architect subcommittees is

\binom{6}{4} = 15

. Adding this to 111 gives

\binom{6+n}{4} = 126

. Since

\binom{9}{4} = 126

, the total pool size is 9 people. Subtracting the 6 architects gives

n = 3

.

Step-by-Step Solution

1

Apply the complementary counting principle

Number of valid subcommittees = (Total subcommittees) - (Subcommittees with 0 urban planners)

It is easier to subtract subcommittees containing only architects from the total number of subcommittees than to sum cases with 1, 2, 3, or 4 urban planners.

2

Calculate the number of subcommittees consisting only of architects

\binom{6}{4} = \frac{6 \times 5}{2 \times 1} = 15

Choosing 4 members from the 6 available architects gives 15 possible all-architect subcommittees.

3

Set up the combination equation for the total pool size

\binom{6+n}{4} - 15 = 111 \implies \binom{6+n}{4} = 126

Adding 15 to both sides isolates the combination expression representing selections from the total pool of

6+n

people.

4

Solve for the total pool size and isolate

n

\binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 \implies 6+n = 9 \implies n = 3

Evaluating combinations shows that selecting 4 items from 9 gives 126. Subtracting the 6 architects leaves 3 urban planners.

Key Concept

Group Selection with Restrictions and Complementary Counting

Estimated Time:2m 0s

Question 16Question →

How many distinct 5-letter code words can be formed by rearranging all of the letters in the word $\text{LEVEL}$ ?

15

24

30

60

120

Show answer & explanation

Answer: $30$

Answer

The total number of distinct 5-letter code words that can be formed is 30.

To calculate the number of unique permutations of a set containing repeated items, divide the factorial of the total number of items by the product of the factorials of the counts of each repeated item. In the word

\text{LEVEL}

, there are 5 letters total, with the letter 'L' appearing 2 times and the letter 'E' appearing 2 times. The calculation is

\frac{5!}{2! \times 2!} = \frac{120}{4} = 30

.

Step-by-Step Solution

1

Identify the total number of letters and the counts of identical letters.

The word

\text{LEVEL}

has 5 total letters, where 'L' appears 2 times and 'E' appears 2 times.

Duplicate elements produce identical arrangements if treated as distinct items.

2

Apply the permutation formula for repeated items:

\frac{n!}{n_1! \times n_2! \times \dots}

.

\frac{5!}{2! \times 2!} = \frac{120}{2 \times 2} = \frac{120}{4} = 30

.

Dividing

5!

by

2!

for the repeated 'L's and

2!

for the repeated 'E's eliminates duplicate arrangements.

Key Concept

Permutations with Repetition / Indistinguishable Objects

Question 17Question →

An integer $n$ is selected at random from the set $\{1, 2, 3, \dots, 500\}$ . What is the probability that $n$ is a perfect square that leaves a remainder of $1$ when divided by $3$ ?

Show answer & explanation

Answer: 0.03

Answer

0.03

The sample space consists of 500 equally likely outcomes. Perfect squares in this range are generated by integers

k

from 1 to 22. Evaluating

k^2 \pmod 3

shows that

k^2

has a remainder of 1 when divided by 3 for all integers

k

not divisible by 3. Excluding the 7 multiples of 3 from the 22 possible values of

k

leaves 15 favorable integers. Dividing the 15 favorable outcomes by the 500 total outcomes gives a probability of 0.03.

Step-by-Step Solution

1

Determine total possible outcomes.

The total number of integers in the set

\{1, 2, 3, \dots, 500\}

is

N = 500

.

Each integer in the set is equally likely to be selected.

2

Find the range of square roots for perfect squares in the set.

Since

22^2 = 484 \le 500

and

23^2 = 529 > 500

,

n = k^2

for

k \in \{1, 2, \dots, 22\}

.

There are 22 perfect squares in the given sample space.

3

Apply remainder constraints to

k^2

.

A square

k^2

leaves a remainder of

1

when divided by

3

if and only if

k

is not divisible by

3

.

If

k

is a multiple of 3,

k^2

is divisible by 3 (remainder 0). If

k

is not a multiple of 3,

k \equiv 1

or

2 \pmod 3

, so

k^2 \equiv 1 \pmod 3

.

4

Count the number of favorable outcomes.

The multiples of 3 up to 22 are

3, 6, 9, 12, 15, 18, 21

(7 numbers). Thus, there are

22 - 7 = 15

values of

k

that satisfy the condition.

Favorable outcomes correspond exactly to integers

k \in \{1, 2, \dots, 22\}

that are not multiples of 3.

5

Compute the single-event probability.

Probability

= \frac{15}{500} = 0.03

.

The probability of a single event is the number of favorable outcomes divided by the total number of outcomes.

Key Concept

Basic Single-Event Probability with Number Properties

Question 18Question →

A computer program randomly selects a single positive integer $x$ from the set $\{1, 2, 3, \dots, 300\}$ , where each integer has an equal probability of being chosen. What is the probability that $x$ is divisible by either 6 or 9, but not by 4?

Show answer & explanation

Answer: 0.14

Answer

The probability that

x

is divisible by 6 or 9, but not by 4, is

0.14

(or

\frac{7}{50}

).

To calculate the probability, determine the ratio of favorable outcomes to total outcomes (300). First, count integers divisible by 6 (50) or 9 (33). Subtracting the 16 multiples of 18 counted in both yields 67 integers. Next, identify how many of these 67 integers are divisible by 4: there are 25 multiples of 12 and 8 multiples of 36 (all 8 of which are already included in the multiples of 12), resulting in 25 integers divisible by 4. Subtracting 25 from 67 leaves 42 favorable integers. The probability is

42 / 300 = 0.14

.

Step-by-Step Solution

1

Determine the total sample space size.

The total number of equally likely outcomes is

|S| = 300

.

Single-event probability requires dividing the count of favorable outcomes by the total number of outcomes in the set.

2

Calculate the count of integers divisible by 6 or 9.

Count =

50 + 33 - 16 = 67

.

Apply the Principle of Inclusion-Exclusion: sum the multiples of 6 and 9, then subtract the multiples of

\text{lcm}(6, 9) = 18

to eliminate double counting.

3

Exclude integers from this set that are divisible by 4.

Count of multiples of 4 to exclude =

25 + 8 - 8 = 25

.

Integers divisible by both 6 and 4 are multiples of

\text{lcm}(6, 4) = 12

. Integers divisible by both 9 and 4 are multiples of

\text{lcm}(9, 4) = 36

. Using inclusion-exclusion,

25 + 8 - 8 = 25

integers must be excluded.

4

Find the number of favorable outcomes and compute the probability.

Favorable outcomes

|E| = 67 - 25 = 42

. Probability

P(E) = \frac{42}{300} = 0.14

.

Divide the favorable set size by the total sample space size.

Key Concept

Basic Single-Event Probability with Set Inclusion-Exclusion and Number Properties

Question 19Question →

A container holds 60 cards, each marked with a distinct integer from 1 to 60, inclusive. If one card is selected at random, the probability that the integer on the selected card is divisible by 5, but divisible by neither 4 nor 6, can be expressed as a simplified fraction $\frac{a}{b}$ , where $a$ and $b$ are positive integers with no common factors greater than 1. What is the value of $a + b$ ?

Show answer & explanation

Answer: 17

Answer

The value of

a + b

is 17.

There are 60 equally likely outcomes. Out of 12 multiples of 5, exactly 4 integers (20, 30, 40, and 60) are divisible by 4 or 6. Thus, 12 - 4 = 8 integers satisfy the condition of being divisible by 5 but neither 4 nor 6. The single-event probability is 8/60 = 2/15. In simplest form, a = 2 and b = 15, yielding a + b = 17.

Step-by-Step Solution

1

Determine total number of possible outcomes.

Total outcomes

N = 60

.

Each of the 60 cards is equally likely to be selected.

2

Identify all multiples of 5 in the set

\{1, 2, \dots, 60\}

.

Multiples of 5:

\{5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60\}

, containing

\frac{60}{5} = 12

integers.

These represent all potential candidates before applying exclusion criteria.

3

Identify multiples of 5 that are also divisible by 4 or by 6.

Multiples of 5 divisible by 4:

\text{LCM}(5,4) = 20 \rightarrow \{20, 40, 60\}

(3 integers). Multiples of 5 divisible by 6:

\text{LCM}(5,6) = 30 \rightarrow \{30, 60\}

(2 integers). Overlap (divisible by both 4 and 6):

\text{LCM}(5,4,6) = 60 \rightarrow \{60\}

(1 integer).

An integer is divisible by both 5 and 4 if it is a multiple of 20, and by both 5 and 6 if it is a multiple of 30.

4

Apply the inclusion-exclusion principle to find the count of invalid multiples of 5.

Invalid multiples =

3 + 2 - 1 = 4

integers (

\{20, 30, 40, 60\}

). Favorable outcomes =

12 - 4 = 8

integers (

\{5, 10, 15, 25, 35, 45, 50, 55\}

).

The integer 60 is divisible by both 4 and 6 and must not be double-subtracted.

5

Calculate the probability, express it as a simplified fraction

\frac{a}{b}

, and compute

a + b

.

Probability

P = \frac{8}{60} = \frac{2}{15}

. Thus,

a = 2

,

b = 15

, and

a + b = 2 + 15 = 17

.

Dividing the numerator and denominator by 4 simplifies

\frac{8}{60}

to

\frac{2}{15}

, where

\text{gcd}(2, 15) = 1

.

Key Concept

Basic Single-Event Probability with Multiple Divisibility Constraints

Question 20Question →

A student must choose $3$ books to read from a reading list of $6$ distinct books. How many different combinations of $3$ books can the student select?

15

18

20

60

120

Show answer & explanation

Answer: $20$

Answer

The student can select

20

different combinations of

3

books.

To find the number of ways to select

3

books from

6

distinct books when order does not matter, use the combination formula

\binom{n}{k} = \frac{n!}{k!(n-k)!}

. Evaluating

\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20

. Thus,

20

is the correct number of combinations.

Step-by-Step Solution

1

Identify the counting method

Since the order in which the books are selected does not matter, use the combinations formula

\binom{n}{k} = \frac{n!}{k!(n-k)!}

.

Selection of a group without regard to order requires combinations.

2

Substitute the given values into the formula

Here

n = 6

and

k = 3

, so

\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1}

.

Apply

n=6

available items and

k=3

chosen items.

3

Simplify the expression

\frac{120}{6} = 20.

Perform the basic arithmetic division.

Key Concept

Combinations and Group Selections

Estimated Time:45s

Counting and Probability

Answer

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