Examkin

Permutations and Combinations

6 soru

Tüm türler6Çoktan seçmeli2Sayısal4
Soru 1Soru →

A project manager must schedule 5 distinct client presentations—for clients A, B, C, D, and E—on 5 consecutive days from Monday through Friday, with exactly one presentation per day. If the presentations for client A and client B cannot be scheduled on consecutive days, how many different presentation schedules are possible?

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Cevap: 72

Cevap

The total number of valid presentation schedules is 72.
The correct result is found by subtracting the number of restricted arrangements (where presentation A and presentation B are scheduled on consecutive days) from the total number of unrestricted arrangements of 5 presentations. The total unrestricted arrangements equal 5!=1205! = 120. Treating A and B as a single block leaves 4 items to arrange in 4!=244! = 24 ways, with 2!=22! = 2 internal orderings for A and B, yielding 24×2=4824 \times 2 = 48 consecutive schedules. Subtracting 48 from 120 results in 72 valid schedules.

Adım Adım Çözüm

1
Calculate total unrestricted linear arrangements of the 5 presentations.
5!=1205! = 120
Without restrictions, 5 distinct items can be arranged in 5 distinct positions in 5!5! ways.
2
Calculate the number of invalid arrangements where presentations A and B are on consecutive days.
2!×4!=482! \times 4! = 48
Grouping A and B into a single unit results in 4 items to arrange (4!=244! = 24), and A and B can swap positions inside the block in 2!=22! = 2 ways.
3
Subtract the invalid arrangements from total arrangements.
120−48=72120 - 48 = 72
Complementary counting gives the total number of arrangements where A and B are not on consecutive days.

Anahtar Kavram

Permutations with Adjacency Restrictions (Complementary Counting)
Soru 2Soru →

A board of directors consists of 77 members. In how many different ways can a subcommittee of 33 members be chosen from the board?

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Cevap: 35

Cevap

35 different subcommittees can be chosen.
The total number of ways to choose a subcommittee of 3 members from a group of 7 members when order does not matter is given by the combination formula (73)=7×6×53×2×1=35\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35.

Adım Adım Çözüm

1
Determine whether the selection requires permutations or combinations.
Since selecting members A, B, and C forms the exact same committee as selecting B, C, and A, the order of selection does not matter. Therefore, this is a combination problem.
Committees are unordered groups.
2
Calculate the number of combinations of 7 items taken 3 at a time using (73)=7!3!(7−3)!\binom{7}{3} = \frac{7!}{3!(7-3)!}.
(73)=7×6×53×2×1=35\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35.
Simplify the factorial expression by canceling out common terms.

Anahtar Kavram

Combinations (Selection Without Regard to Order)
Soru 3Soru →

A financial advisory board of 66 members is to be selected from a pool of 66 senior analysts and 55 junior analysts. The board must include at least 22 senior analysts and at least 22 junior analysts. However, two specific senior analysts, AA and BB, cannot both serve on the board together. How many different 66-member boards can be formed under these conditions?

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Cevap: 320

Cevap

The total number of valid 6-member boards that can be formed is 320.
To find the number of valid boards, we first calculate all possible 6-member boards satisfying the minimum criteria of having at least 2 senior analysts and at least 2 junior analysts. The possible (senior, junior) distributions are (4,2), (3,3), and (2,4). Calculating each case gives (64)(52)=150\binom{6}{4}\binom{5}{2} = 150, (63)(53)=200\binom{6}{3}\binom{5}{3} = 200, and (62)(54)=75\binom{6}{2}\binom{5}{4} = 75, for a total of 150+200+75=425150 + 200 + 75 = 425 boards. Next, we determine how many of these boards contain both senior analysts A and B. Fixing A and B requires choosing 4 more members from the remaining 4 seniors and 5 juniors such that the total junior count is at least 2. The valid remaining senior choices s′s' are 0, 1, or 2, yielding (40)(54)=5\binom{4}{0}\binom{5}{4} = 5, (41)(53)=40\binom{4}{1}\binom{5}{3} = 40, and (42)(52)=60\binom{4}{2}\binom{5}{2} = 60, totaling 5+40+60=1055 + 40 + 60 = 105 boards. Subtracting these forbidden boards from 425 yields 425−105=320425 - 105 = 320.

Adım Adım Çözüm

1
Calculate total valid boards with at least 2 senior analysts and at least 2 junior analysts without restriction on A and B.
Total valid boards without conflict constraint = 425
The valid (Senior, Junior) distributions for a 6-member board are (4,2), (3,3), and (2,4). Their combinations are (64)(52)=150\binom{6}{4}\binom{5}{2} = 150, (63)(53)=200\binom{6}{3}\binom{5}{3} = 200, and (62)(54)=75\binom{6}{2}\binom{5}{4} = 75, giving 150+200+75=425150 + 200 + 75 = 425.
2
Calculate the number of valid boards that contain BOTH senior analysts A and B.
Number of invalid boards containing both A and B = 105
If A and B are both selected, 2 senior spots are fixed. We need 4 remaining members from the remaining 4 seniors and 5 juniors. To satisfy the junior constraint (at least 2 juniors), the additional senior count s′s' can be 0, 1, or 2: s′=0  ⟹  (40)(54)=5s'=0 \implies \binom{4}{0}\binom{5}{4}=5, s′=1  ⟹  (41)(53)=40s'=1 \implies \binom{4}{1}\binom{5}{3}=40, s′=2  ⟹  (42)(52)=60s'=2 \implies \binom{4}{2}\binom{5}{2}=60. Summing gives 5+40+60=1055 + 40 + 60 = 105.
3
Subtract the conflicting boards containing both A and B from the total valid boards.
Valid boards = 425 - 105 = 320
Complementary counting gives the exact number of boards satisfying all conditions.

Anahtar Kavram

Combinations with Subgroup Constraints and Complementary Counting
Tahmini Süre:2m 0s
Soru 4Soru →

A 5-digit passcode is to be created using five distinct digits chosen from the set {1,2,3,4,5,6,7}\{1, 2, 3, 4, 5, 6, 7\}. The passcode must include both the digit 33 and the digit 55, and the digit 33 must appear somewhere to the left of the digit 55 in the passcode. How many such 5-digit passcodes can be formed?

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Cevap: 600

Cevap

600
To form a valid passcode, first select 3 digits from the 5 available digits {1,2,4,6,7}\{1, 2, 4, 6, 7\}, which can be done in (53)=10\binom{5}{3} = 10 ways. Each chosen set of 5 distinct digits (including 3 and 5) can be arranged in 5!=1205! = 120 ways. Because the digits 3 and 5 are distinct, the digit 3 appears before the digit 5 in exactly half of these arrangements (120/2=60120 / 2 = 60). Multiplying the 10 combinations of digits by the 60 valid arrangements gives a total of 600600 valid passcodes.

Adım Adım Çözüm

1
Determine the number of ways to choose the remaining digits.
(53)=10\binom{5}{3} = 10 ways to choose 3 additional digits from {1,2,4,6,7}\{1, 2, 4, 6, 7\}.
Since the passcode must contain both 3 and 5, 3 additional distinct digits must be selected from the 5 available remaining digits.
2
Calculate the total permutations of the 5 chosen digits.
5!=1205! = 120 total permutations.
Any set of 5 distinct digits can be arranged into a 5-digit sequence in 5!5! ways.
3
Apply the positional restriction using symmetry.
1202=60\frac{120}{2} = 60 valid arrangements per set of digits.
In exactly half of all permutations containing both 3 and 5, the digit 3 appears before the digit 5.
4
Compute the total number of valid passcodes.
10×60=60010 \times 60 = 600 passcodes.
Multiply the number of digit selections by the number of valid orderings per selection.

Anahtar Kavram

Combining selection (combinations) with symmetry-restricted arrangements (permutations)
Tahmini Süre:2m 0s
Soru 5Soru →

A security system generates 4-digit passcodes using distinct digits selected from the set {1,2,3,4,5,6,7}\{1, 2, 3, 4, 5, 6, 7\}. How many different 4-digit passcodes can be formed if the first digit must be odd and the last digit must be even?

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Cevap: 240

Cevap

240
To form a valid passcode under the given constraints, we analyze each position step-by-step. The first digit has 4 possible choices (odd numbers: 1, 3, 5, 7), and the fourth digit has 3 possible choices (even numbers: 2, 4, 6). Since the problem specifies that the digits in the passcode must be distinct, choosing the first and last digits consumes 2 of the 7 available digits, leaving 5 digits for the second position and 4 digits for the third position. Applying the Fundamental Counting Principle gives a total of 4×5×4×3=2404 \times 5 \times 4 \times 3 = 240 unique passcodes.

Adım Adım Çözüm

1
Determine choices for the first digit
4 choices
The set contains 4 odd digits: 1, 3, 5, and 7.
2
Determine choices for the fourth (last) digit
3 choices
The set contains 3 even digits: 2, 4, and 6.
3
Determine choices for the second and third digits
5 choices for the second digit, 4 choices for the third digit
All digits in the passcode must be distinct. Having used 2 digits for the first and last positions, 5 digits remain out of 7 for the second slot, and 4 digits remain for the third slot.
4
Apply the Fundamental Counting Principle
4 × 5 × 4 × 3 = 240
Multiply the number of available options for each position to find the total number of distinct passcodes.

Anahtar Kavram

Permutations with Position Restrictions
Soru 6Soru →

A panel discussion features 6 participants: 3 scientists, 2 economists, and 1 moderator. In how many different linear seating arrangements can these 6 participants be seated in a single row of 6 chairs if all 3 scientists must sit in adjacent seats?

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Cevap: 144144

Cevap

The total number of valid seating arrangements is 144.
To arrange participants under the condition that all 3 scientists sit together, treat the 3 scientists as a single block. This leaves 4 units to arrange: 1 block of scientists, 2 economists, and 1 moderator. These 4 units can be arranged in a row in 4!=244! = 24 ways. Furthermore, within the scientist block, the 3 individual scientists can be arranged in 3!=63! = 6 distinct orders. By the fundamental counting principle, multiplying the arrangements of the main units by the internal arrangements of the block gives 24×6=14424 \times 6 = 144 total valid seating arrangements.

Adım Adım Çözüm

1
Group the restricted items into a single block.
Treat the 3 scientists as 1 single block unit.
Because the 3 scientists must sit together in adjacent seats, they move as a single block along with the other participants.
2
Calculate the number of arrangements for the distinct units.
There are 4 units to arrange (1 scientist block + 2 economists + 1 moderator), giving 4!=244! = 24 arrangements.
The Fundamental Counting Principle states that nn distinct items can be linearly arranged in n!n! ways.
3
Calculate the internal arrangements within the block.
The 3 scientists can be arranged among themselves within their block in 3!=63! = 6 ways.
The order of the individual scientists inside the block matters.
4
Multiply the unit arrangements by the internal arrangements.
Total arrangements = 4!×3!=24×6=1444! \times 3! = 24 \times 6 = 144.
By the multiplication principle, the total ways to complete both independent arrangement tasks is the product of their individual possibilities.

Anahtar Kavram

Permutations with Adjacency Restrictions (Block Method)
Tahmini Süre:1m 30s
Permutations and Combinations Alıştırma Soruları — GMAT | Examkin