Counting and Probability

91 soru

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An artisan tea shop allows customers to order a custom tea blend prepared through three sequential choices:

1. Base Tea: Choose $1$ of $4$ available tea types (Black, Green, Oolong, or White).
2. Flavor Infusion: Choose $1$ of $5$ available herbal flavors. However, if Black tea is chosen as the base, only $3$ of these $5$ herbal flavors can be selected.
3. Sweetener: Choose $1$ of $3$ natural sweeteners (Honey, Stevia, or Maple Syrup), or choose to have no sweetener.

How many different custom tea blends can a customer create?

54

60

72

80

96

Cevabı ve açıklamayı göster

Cevap: $72$

Cevap

72 different custom tea blends can be created.

To solve this counting problem with conditional restrictions, break the calculation into two mutually exclusive cases based on the base tea chosen:

1. Case 1: Black Tea Base
- Base tea options:

1

(Black)
- Flavor infusion options:

3

(restricted subset)
- Sweetener options:

4

(

3

sweeteners +

1

option for no sweetener)
- Combinations =

1 \times 3 \times 4 = 12

2. Case 2: Non-Black Tea Base (Green, Oolong, or White)
- Base tea options:

3

- Flavor infusion options:

5

(all available)
- Sweetener options:

4

- Combinations =

3 \times 5 \times 4 = 60

Adding the combinations from these two mutually exclusive cases yields

12 + 60 = 72

total custom tea blends.

Adım Adım Çözüm

Identify total options available for the sweetener stage.

There are

3

sweetener choices plus

1

choice for 'no sweetener', making

3 + 1 = 4

total options for Stage 3.

Choosing 'no sweetener' is a distinct decision outcome that must be counted.

Calculate the number of blends when Black tea is chosen as the base.

Number of Black tea blends =

1 \text{ (Black base)} \times 3 \text{ (flavors)} \times 4 \text{ (sweetener choices)} = 12

When Black tea is chosen, only

3

flavor infusions are permitted.

Calculate the number of blends when any non-Black tea is chosen as the base.

Number of non-Black tea blends =

3 \text{ (Green, Oolong, White bases)} \times 5 \text{ (flavors)} \times 4 \text{ (sweetener choices)} = 60

There are

3

non-Black base choices, and each can be paired with any of the

5

flavor infusions.

Sum the combinations from both mutually exclusive cases.

Total blends =

12 + 60 = 72

Since selecting Black tea and selecting non-Black tea are mutually exclusive cases, their individual counts are added.

Anahtar Kavram

Fundamental Counting Principle with Conditional Restrictions

Tahmini Süre:1m 45s

Soru 82Soru →

A jewelry store window displays $6$ distinct luxury watches— $3$ gold watches and $3$ silver watches—in a single straight line from left to right. If no two gold watches can be placed adjacent to each other, how many different linear arrangements of the $6$ watches are possible?

Cevabı ve açıklamayı göster

Cevap: 144

Cevap

144

To place items such that no two gold watches are adjacent, we use the gap method. First, arrange the 3 distinct silver watches, which can be done in

3! = 6

ways. Placing these 3 silver watches creates 4 potential spaces (one at each end and two between the silver watches). Next, select 3 of these 4 spaces to place the gold watches, which can be done in

\binom{4}{3} = 4

ways. Finally, arrange the 3 distinct gold watches within the selected spaces in

3! = 6

ways. Multiplying these possibilities gives

6 \times 4 \times 6 = 144

total distinct arrangements.

Adım Adım Çözüm

Calculate the arrangements of the 3 distinct silver watches.

3! = 6 ways

The 3 silver watches are distinct and can be arranged among themselves in 3! ways.

Determine the available positions (gaps) for the gold watches.

4 available gaps

Placing 3 silver watches in a line creates 4 potential spaces (before the first, between adjacent pairs, and after the last) to ensure non-adjacency.

Select positions and arrange the 3 distinct gold watches.

C(4, 3) × 3! = 4 × 6 = 24 ways

Choosing 3 out of 4 gaps gives C(4, 3) = 4 combinations, and ordering the 3 distinct gold watches in those selected gaps gives 3! = 6 arrangements.

Multiply the possibilities from all independent steps using the Fundamental Counting Principle.

6 × 24 = 144

The total number of valid linear arrangements is the product of the number of ways to complete each step.

Anahtar Kavram

Linear Permutations with Non-Adjacent Restrictions (Gap Method)

Tahmini Süre:2m 0s

Soru 83Soru →

A corporate board of 7 members—consisting of 4 senior executives (including the CEO and the COO) and 3 junior associates—is to be seated in a single row of 7 chairs for a press conference. If the 3 junior associates must sit in adjacent chairs, and the CEO and the COO cannot sit next to each other, in how many different linear arrangements can the 7 members be seated?

288

432

576

720

1,440

Cevabı ve açıklamayı göster

Cevap: 432

Cevap

432

To satisfy the condition that the 3 junior associates sit together, we treat them as 1 block with

3! = 6

internal orderings. Combining this block with the 4 senior executives yields 5 entities, which can be arranged in

5! = 120

ways, giving

120 \times 6 = 720

total arrangements with the juniors seated together. To enforce that the CEO and COO cannot sit together, we subtract the arrangements where they do sit together: treating the CEO and COO as a block gives 4 entities to arrange (

4! = 24

), with

2! = 2

ways to arrange CEO and COO, and

3! = 6

ways for the junior block, totaling

24 \times 2 \times 6 = 288

restricted cases. Subtracting 288 from 720 gives 432 valid arrangements.

Adım Adım Çözüm

Group the 3 junior associates into a single block.

The 3 junior associates can be arranged internally within their block in

3! = 6

ways. Treating this block as 1 single element along with the 4 senior executives gives a total of 5 items to arrange.

The condition specifies that all 3 junior associates must sit in adjacent chairs.

Calculate total arrangements where junior associates sit together without CEO/COO restrictions.

The 5 items (1 block + 4 senior executives) can be arranged in

5! = 120

ways. Including internal block arrangements gives

120 \times 6 = 720

ways.

Applying the Fundamental Counting Principle to the 5 units and the 3 internal positions.

Calculate unwanted arrangements where the CEO and COO sit next to each other (with junior associates together).

Group the CEO and COO into a second block with

2! = 2

internal arrangements. Now there are 4 items to arrange (Junior block, CEO-COO block, and 2 other senior executives). Total unwanted arrangements:

4! \times 3! \times 2! = 24 \times 6 \times 2 = 288

ways.

To find valid non-adjacent arrangements, subtract adjacent CEO-COO arrangements from total junior-grouped arrangements.

Subtract the unwanted arrangements from the total grouped arrangements.

720 - 288 = 432

valid seating arrangements.

Complementary counting provides a direct solution.

Anahtar Kavram

Linear Permutations with Grouping and Complementary Non-Adjacency Constraints

Soru 84Soru →

An architectural firm offers custom townhouse design packages created by making sequential choices across three stages: Exterior Facade Material (4 choices), Floor Plan Layout (5 choices), and Interior Lighting Scheme (6 choices).

However, two design constraints apply:
1. If the first exterior facade material option is chosen, the fifth floor plan layout option cannot be selected.
2. The sixth interior lighting scheme option can only be selected if the fifth floor plan layout option is chosen.

How many different valid townhouse design packages can a client create?

Cevabı ve açıklamayı göster

Cevap: 98

Cevap

The total number of valid townhouse design packages is 98.

The solution splits the decision process into two mutually exclusive scenarios based on whether the 5th layout option is chosen. In Case 1 (5th layout chosen), 3 facade choices × 1 layout choice × 6 lighting choices yields 18 valid packages. In Case 2 (1st through 4th layout chosen), 4 facade choices × 4 layout choices × 5 lighting choices yields 80 valid packages. Adding these mutually exclusive totals gives 18 + 80 = 98 valid design packages.

Adım Adım Çözüm

Analyze the decision tree by splitting into mutually exclusive cases based on whether Floor Plan Layout Option 5 is selected.

Two distinct cases are established: Case 1 (Layout Option 5 selected) and Case 2 (Layout Options 1–4 selected).

The constraints depend directly on whether Floor Plan Layout Option 5 is chosen.

Calculate valid combinations for Case 1 (Layout Option 5 selected).

Facade choices = 3 (excluding Option 1), Layout choices = 1 (Option 5), Lighting choices = 6 (including Option 6). Total = 3 × 1 × 6 = 18 packages.

Constraint 1 eliminates 1 facade option, while Constraint 2 allows all 6 lighting options.

Calculate valid combinations for Case 2 (Layout Option 5 NOT selected).

Facade choices = 4, Layout choices = 4 (Options 1–4), Lighting choices = 5 (excluding Option 6). Total = 4 × 4 × 5 = 80 packages.

Constraint 1 does not apply, but Constraint 2 prevents selecting Lighting Option 6.

Sum the valid outcomes from both mutually exclusive cases.

18 + 80 = 98 valid packages.

By the Addition Principle of Counting, non-overlapping cases are added together.

Anahtar Kavram

Fundamental Counting Principle with Constrained Cases

Soru 85Soru →

A secure access code is formed by arranging all $6$ of the distinct characters $A, E, B, C, D,$ and $F$ in a single line. If the two vowels ( $A$ and $E$ ) cannot be adjacent to each other, and the code cannot begin with the consonant $B$ , how many different valid access codes can be formed?

240

252

360

408

480

Cevabı ve açıklamayı göster

Cevap: 408

Cevap

408 valid access codes can be formed.

To find the number of valid arrangements satisfying both conditions, we use complementary counting. First, the total number of arrangements of 6 distinct characters is 6! = 720. Next, we determine the number of invalid arrangements. Let V be the set of arrangements where A and E are adjacent: treating (AE) as a single unit gives 5! × 2! = 240 ways. Let B_first be the set of arrangements starting with B: fixing B in the first spot leaves 5! = 120 ways. The intersection of these two invalid sets (where A and E are adjacent AND the code starts with B) places B in position 1 and arranges the remaining 4 items including the (AE) block in 4! × 2! = 48 ways. By inclusion-exclusion, the total invalid arrangements equal 240 + 120 - 48 = 312. Subtracting invalid arrangements from the total gives 720 - 312 = 408 valid access codes.

Adım Adım Çözüm

Calculate the total total number of unrestricted linear arrangements of the 6 distinct characters.

6! = 720 total arrangements.

6 distinct characters can be ordered in 6! different ways.

Calculate the number of arrangements where the two vowels (A and E) ARE adjacent (Set V).

|V| = 240 arrangements.

Treating (AE) as 1 block leaves 5 items to arrange in 5! = 120 ways. Since (AE) has 2! = 2 internal arrangements, |V| = 2 * 120 = 240.

Calculate the number of arrangements that BEGIN with the consonant B (Set B).

|B| = 120 arrangements.

Fixing B in the first position leaves 5 remaining positions to be filled by the remaining 5 characters in 5! = 120 ways.

Calculate the overlap: arrangements where vowels ARE adjacent AND the code BEGINS with B (Set V ∩ B).

|V ∩ B| = 48 arrangements.

With B fixed in position 1, the remaining 5 positions are filled by 4 blocks: (AE), C, D, F. These 4 blocks can be arranged in 4! = 24 ways, and (AE) has 2! = 2 internal orderings, giving 24 * 2 = 48.

Apply the Principle of Inclusion-Exclusion to find the total invalid arrangements, and subtract from the total.

720 - (240 + 120 - 48) = 720 - 312 = 408 valid arrangements.

The number of invalid arrangements is |V ∪ B| = |V| + |B| - |V ∩ B| = 312. Subtracting invalid codes from total codes yields 408.

Anahtar Kavram

Linear permutations with multiple restrictions using complementary counting and inclusion-exclusion.

Tahmini Süre:2m 0s

Soru 86Soru →

An event organizer is scheduling a conference featuring $7$ distinct speakers: $3$ technology experts, $3$ finance experts, and $1$ keynote presenter. The $7$ speakers will give back-to-back presentations on a single main stage. If all $3$ technology experts must speak consecutively in any order, and no two finance experts may speak consecutively, how many different presentation schedules are possible?

Cevabı ve açıklamayı göster

Cevap: 72

Cevap

The total number of possible presentation schedules satisfying all conditions is 72.

To find the number of valid schedules, we combine the block method and the gap method. First, bundle the 3 technology experts into a single block. This block can be ordered internally in

3! = 6

ways. Next, arrange the non-finance items (the technology block and the keynote presenter), which can be ordered in

2! = 2

ways. These 2 items create 3 available gaps (one at each end and one in the middle) for the 3 finance experts. To ensure no two finance experts are consecutive, place exactly one finance expert into each of the 3 gaps, which can be done in

3! = 6

ways. Multiplying these independent choices yields

6 \times 2 \times 6 = 72

total valid schedules.

Adım Adım Çözüm

Group the 3 technology experts into a single unit and calculate internal permutations

3! = 6 internal arrangements for the technology block

The technology experts must appear consecutively, so treating them as a single block ensures they remain together.

Arrange the non-finance elements (the technology block and the keynote presenter)

2! = 2 linear arrangements

There are 2 distinct items that act as separators for the finance experts.

Apply the gap method to place the 3 finance experts into non-adjacent slots

3 available gaps for 3 finance experts gives 3! = 6 ways

Placing at most one finance expert in each gap created by the non-finance items guarantees that no two finance experts are seated or scheduled consecutively.

Calculate total permutations using the Fundamental Counting Principle

6 × 2 × 6 = 72 total schedules

The decisions for internal block order, base item order, and gap assignments are independent.

Anahtar Kavram

Permutations with Combined Block and Separation (Gap Method) Restrictions

Soru 87Soru →

A restaurant culinary team is designing a tasting menu consisting of $4$ distinct dishes selected from a pool of $5$ appetizers and $6$ main courses. The tasting menu must contain at least $1$ appetizer and at least $2$ main courses. Furthermore, a specific appetizer, Dish A, can only be included on the menu if a specific main course, Dish B, is also included. How many different valid tasting menus can be formed?

180

200

240

250

310

Cevabı ve açıklamayı göster

Cevap: 200

Cevap

200 different valid tasting menus can be formed.

The total number of valid tasting menus is

200

. This is calculated by breaking down the total requirement into two mutually exclusive dish distributions: 1 appetizer with 3 main courses (giving

90

valid menus) and 2 appetizers with 2 main courses (giving

110

valid menus), accounting for the restriction that selecting Dish A requires Dish B.

Adım Adım Çözüm

Determine the valid distribution of dishes between appetizers and main courses.

Since

4

dishes must be selected with at least

1

appetizer and at least

2

main courses, the only possible distributions are: Case 1 (

1

appetizer,

3

main courses) and Case 2 (

2

appetizers,

2

main courses).

Selecting

3

appetizers and

1

main course violates the constraint of having at least

2

main courses.

Calculate the valid combinations for Case 1 (1 appetizer and 3 main courses).

Subcase 1a (Dish A not selected): Choose

1

appetizer from the remaining

4

appetizers and

3

main courses from all

6

main courses:

\binom{4}{1} \times \binom{6}{3} = 4 \times 20 = 80

.
Subcase 1b (Dish A selected): Dish B must also be selected. Choose no additional appetizers from the remaining

4

, and choose

2

more main courses from the remaining

5

main courses:

1 \times \binom{5}{2} = 10

.
Total for Case 1:

80 + 10 = 90

Separating into cases based on whether Dish A is included properly enforces the conditional constraint.

Calculate the valid combinations for Case 2 (2 appetizers and 2 main courses).

Subcase 2a (Dish A not selected): Choose

2

appetizers from the remaining

4

appetizers and

2

main courses from all

6

main courses:

\binom{4}{2} \times \binom{6}{2} = 6 \times 15 = 90

.
Subcase 2b (Dish A selected): Dish B must also be selected. Choose

1

more appetizer from the remaining

4

appetizers and

1

more main course from the remaining

5

main courses:

\binom{4}{1} \times \binom{5}{1} = 4 \times 5 = 20

.
Total for Case 2:

90 + 20 = 110

Evaluates all valid menus having

2

appetizers and

2

main courses.

Sum the valid combinations from both cases.

Total valid menus =

90 + 110 = 200

The two main distribution cases are mutually exclusive.

Anahtar Kavram

Combinations with multiple category constraints and conditional dependencies

Soru 88Soru →

A university research department must select a 4-member committee from a group of 6 biologists and 4 chemists. If the committee must contain at least 1 biologist and at least 1 chemist, how many different 4-member committees can be formed?

Cevabı ve açıklamayı göster

Cevap: 194

Cevap

194

Using complementary counting, the total number of ways to pick 4 members out of 10 is 210. The scenarios that break the rule are committees with all biologists (15 ways) or all chemists (1 way). Subtracting these 16 restricted selections from 210 results in 194 valid committees.

Adım Adım Çözüm

Find total unrestricted combinations

210 total committees

Choosing any 4 people out of the pool of 10 without restrictions yields 10C4 = 210.

Calculate restricted/invalid cases

16 invalid committees

All-biologist groups account for 6C4 = 15 combinations, and all-chemist groups account for 4C4 = 1 combination.

Apply complementary counting

194 valid committees

Subtracting the 16 invalid combinations from the total 210 combinations gives 194.

Anahtar Kavram

Combinations with Restrictions (Complementary Counting)

Soru 89Soru →

A technology startup needs to form a 4-member project team selected from a pool of 5 front-end developers and 5 back-end developers. However, the selection is subject to one condition: if a specific front-end developer, Elena, is selected for the team, then a specific back-end developer, Marcus, must also be selected for the team. How many different 4-member teams can be formed under this condition?

Cevabı ve açıklamayı göster

Cevap: 154

Cevap

154 different 4-member teams can be formed under the given condition.

The total number of ways to select a 4-member team from 10 developers without restrictions is

\binom{10}{4} = 210

. The conditional restriction requires that whenever Elena is on the team, Marcus must also be on the team. The only selection that violates this rule is one where Elena is included but Marcus is excluded. For this invalid scenario, Elena takes 1 spot and Marcus is removed from consideration, leaving

\binom{8}{3} = 56

ways to pick the remaining 3 members from the remaining 8 developers. Subtracting these 56 invalid teams from the 210 total teams gives 154 valid teams.

Adım Adım Çözüm

Calculate the total possible 4-member teams from the 10 developers without any restrictions.

Total combinations =

\binom{10}{4} = 210

Choosing any 4 members out of a total pool of 10 developers (5 front-end and 5 back-end).

Identify and calculate the number of invalid team combinations.

Invalid combinations =

\binom{8}{3} = 56

The conditional rule is violated only if Elena is selected while Marcus is excluded. Selecting Elena fills 1 spot and excluding Marcus leaves 8 developers from which to choose the remaining 3 team members.

Subtract the invalid combinations from the total unconstrained combinations.

210 - 56 = 154

Using complementary counting yields the number of valid team selections directly and efficiently.

Anahtar Kavram

Combinations with Conditional Restrictions (Complementary Counting)

Tahmini Süre:1m 30s

Soru 90Soru →

A committee of $4$ delegates is to be selected from a group of $5$ diplomats from Country X and $4$ diplomats from Country Y. If $1$ specific diplomat from Country X and $1$ specific diplomat from Country Y cannot both serve on the committee together, how many different $4$ -member committees can be formed?

105

126

Cevabı ve açıklamayı göster

Cevap: 105

Cevap

105

To find the number of valid 4-member committees, calculate the total combinations of choosing 4 delegates from 9 diplomats,

\binom{9}{4} = 126

. Then subtract the restricted outcomes where both specific diplomats are selected. If both specific diplomats are included, the remaining 2 positions on the committee must be chosen from the remaining 7 diplomats, which can be done in

\binom{7}{2} = 21

ways. Subtracting these invalid outcomes gives

126 - 21 = 105

Adım Adım Çözüm

Calculate the total number of ways to choose 4 delegates from the total pool of 9 diplomats without any restrictions.

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

The total pool consists of 5 diplomats from Country X + 4 diplomats from Country Y = 9 diplomats.

Calculate the number of invalid committee selections where both specific diplomats are selected together.

\binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21

If the 2 specific diplomats are both placed on the committee, 2 remaining seats must be filled from the remaining 7 diplomats.

Subtract the invalid combinations from the total combinations to find the number of valid committees.

126 - 21 = 105

Complementary counting gives the number of committees where both specific diplomats do not serve together.

Anahtar Kavram

Combinations with restrictions using complementary counting

Tahmini Süre:2m 0s

Soru 91Soru →

A cybersecurity firm must select a 4-tool diagnostic suite from a pool of 6 network scanners and 4 code analyzers. If the suite must contain at least 1 network scanner and at least 1 code analyzer, how many different 4-tool suites can be selected?

Cevabı ve açıklamayı göster

Cevap: 194

Cevap

The total number of different 4-tool suites that can be selected is 194.

The total number of ways to select any 4 tools from the pool of 10 is

\binom{10}{4} = 210

. The only selections that violate the requirement of having at least one tool of each type are choosing 4 scanners and 0 analyzers (

\binom{6}{4} = 15

ways) or 0 scanners and 4 analyzers (

\binom{4}{4} = 1

way). Subtracting these 16 invalid outcomes from the total gives

210 - 16 = 194

valid 4-tool suites.

Adım Adım Çözüm

Calculate the total unconstrained selections

210

There are

6 + 4 = 10

total tools, and 4 must be chosen.

\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210

Determine the number of invalid combinations violating the constraint

16 invalid combinations

The constraint requires at least 1 scanner and 1 analyzer. Invalid suites consist entirely of scanners (

\binom{6}{4} = 15

) or entirely of analyzers (

\binom{4}{4} = 1

Apply complementary counting to find the valid combinations

194 valid combinations

Subtract the invalid combinations from total unconstrained combinations:

210 - (15 + 1) = 194

Anahtar Kavram

Combinations with Group Selections and Complementary Counting

ÖncekiSayfa 5 / 5

Counting and Probability

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