Examkin

Permutations

15 soru

Tüm türler15Çoktan seçmeli7Sayısal8
Soru 1Soru →

In how many different ways can a chairperson and a secretary be selected from a committee of 55 members?

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Cevap: 2020

Cevap

The number of ways to select a chairperson and a secretary from 55 members is 2020.
Selecting 22 individuals for distinct positions (chairperson and secretary) from 55 candidates is an ordered selection problem. The number of ways is given by the permutation formula 5P2=5×4=20_5P_2 = 5 \times 4 = 20.

Adım Adım Çözüm

1
Identify whether order matters
Order matters because the roles of chairperson and secretary are distinct.
When distinct roles are assigned, the arrangement is a permutation rather than a combination.
2
Apply the permutation formula nPr=n!(n−r)!_nP_r = \frac{n!}{(n-r)!} for n=5n=5 and r=2r=2
5P2=5!(5−2)!=5!3!=5×4=20_5P_2 = \frac{5!}{(5-2)!} = \frac{5!}{3!} = 5 \times 4 = 20
There are 55 choices for chairperson and 44 remaining choices for secretary.

Anahtar Kavram

Permutation of nn distinct items taken rr at a time
Tahmini Süre:45s
Soru 2Soru →

In how many different ways can the letters of the word SUCCESS be arranged such that the three 'S's do not all come together?

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Cevap: 360

Cevap

The letters of the word SUCCESS can be arranged in 360 ways such that the three 'S's do not all come together.
The correct answer is calculated using complementary counting. First, the total unrestricted permutations of SUCCESS (7 letters with 3 'S's and 2 'C's) is 7!3!2!=420\frac{7!}{3!2!} = 420. Next, treating the three 'S's as one single block leaves 5 items to arrange with 2 'C's, giving 5!2!=60\frac{5!}{2!} = 60 ways where the 'S's are together. Subtracting 60 from 420 yields 360.

Adım Adım Çözüm

1
Calculate the total number of unrestricted arrangements of the word SUCCESS.
The word SUCCESS has 7 letters in total: 3 'S's, 2 'C's, 1 'U', and 1 'E'. Total arrangements Ntotal=7!3!×2!=50406×2=420N_{total} = \frac{7!}{3! \times 2!} = \frac{5040}{6 \times 2} = 420.
Repeated letters must be accounted for by dividing the factorial of the total count by the factorials of the counts of repeated letters.
2
Calculate the number of arrangements where the three 'S's are all together.
Treat the three 'S's as a single entity (SSS). We now arrange 5 entities: (SSS), U, C, C, E. Since 'C' appears twice, Ntogether=5!2!=1202=60N_{together} = \frac{5!}{2!} = \frac{120}{2} = 60.
Grouping restricted identical items into a single block allows us to find the subset of arrangements where they stay together.
3
Subtract the number of 'together' arrangements from the total arrangements.
Nnot_together=Ntotal−Ntogether=420−60=360N_{not\_together} = N_{total} - N_{together} = 420 - 60 = 360.
The complementary counting principle gives the number of ways where the restriction is satisfied.

Anahtar Kavram

Permutations with Repeated Elements and Complementary Counting
Tahmini Süre:1m 30s
Soru 3Soru →

Find the positive integer value of nn such that nP4=42×nP2^{n}P_4 = 42 \times {^{n}P_2}.

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Cevap: 9

Cevap

The positive integer value of nn is 9.
Expanding nP4^{n}P_4 as n(n−1)(n−2)(n−3)n(n-1)(n-2)(n-3) and nP2^{n}P_2 as n(n−1)n(n-1) allows dividing out n(n−1)n(n-1), leading to (n−2)(n−3)=42(n-2)(n-3) = 42. Expanding and factoring gives n2−5n−36=0n^2 - 5n - 36 = 0, which yields n=9n = 9 as the only valid positive integer.

Adım Adım Çözüm

1
Apply the permutation formula nPr=n!(n−r)!^{n}P_r = \frac{n!}{(n-r)!}
nP4=n(n−1)(n−2)(n−3)^{n}P_4 = n(n-1)(n-2)(n-3) and nP2=n(n−1)^{n}P_2 = n(n-1)
By definition of permutations, selecting rr items from nn distinct items without replacement.
2
Substitute the expansions into the given relation
n(n−1)(n−2)(n−3)=42n(n−1)n(n-1)(n-2)(n-3) = 42 n(n-1)
Direct substitution into nP4=42×nP2^{n}P_4 = 42 \times {^{n}P_2}.
3
Simplify by dividing out common non-zero terms
(n−2)(n−3)=42(n-2)(n-3) = 42
Since n≥4n \ge 4, n(n−1)≠0n(n-1) \neq 0 and can be safely divided from both sides.
4
Form and solve the quadratic equation
n2−5n+6=42  ⟹  n2−5n−36=0  ⟹  (n−9)(n+4)=0n^2 - 5n + 6 = 42 \implies n^2 - 5n - 36 = 0 \implies (n-9)(n+4) = 0
Expanding terms and factoring the resulting quadratic expression.
5
Determine the valid root
n=9n = 9
Permutation total items nn must satisfy n≥r≥0n \ge r \ge 0, rejecting the negative root n=−4n = -4.

Anahtar Kavram

Algebraic equations involving permutations
Tahmini Süre:1m 15s
Soru 4Soru →

Six members of a board of directors are to be seated around a circular conference table. If two specific members refuse to sit next to each other, how many different seating arrangements are possible?

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Cevap: 72

Cevap

72
To find the number of circular seating arrangements where two specific members do not sit next to each other, use the complement rule. First, the total unrestricted circular arrangements of 6 members is (6−1)!=5!=120(6 - 1)! = 5! = 120. Next, calculate the arrangements where the two members sit together by treating them as 1 unit (giving 5 units in total). The circular arrangements of these 5 units is (5−1)!=4!=24(5 - 1)! = 4! = 24, and the 2 members can swap seats in 2!=22! = 2 ways, yielding 24×2=4824 \times 2 = 48 arrangements together. Subtracting this from the total yields 120−48=72120 - 48 = 72.

Adım Adım Çözüm

1
Calculate total circular arrangements without restrictions
Total arrangements = (6−1)!=5!=120(6 - 1)! = 5! = 120
The number of ways to arrange nn distinct items in a circle is (n−1)!(n - 1)!.
2
Calculate arrangements where the two specific members sit together
Restricted arrangements = (5−1)!×2!=4!×2=24×2=48(5 - 1)! \times 2! = 4! \times 2 = 24 \times 2 = 48
Treat the two members as a single block, giving 5 units to arrange around a circle in (5−1)!(5-1)! ways, and multiply by 2!2! for internal ordering of the pair.
3
Subtract the together arrangements from total arrangements
Ways apart = 120−48=72120 - 48 = 72
Complementary counting gives the number of arrangements where the two members do not sit next to each other.

Anahtar Kavram

Circular Permutations with Restrictions
Tahmini Süre:1m 30s
Soru 5Soru →

In how many different ways can 55 boys and 33 girls be seated in a straight row such that all 33 girls must sit together?

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Cevap: 4320

Cevap

The total number of ways to seat 55 boys and 33 girls in a row such that all 33 girls sit together is 43204320.
To arrange 55 boys and 33 girls so that the girls are always together, treat the 33 girls as 11 single unit. Combined with the 55 boys, there are 66 units to arrange in a straight line, which can be done in 6!=7206! = 720 ways. Within their group, the 33 girls can be arranged in 3!=63! = 6 ways. By the multiplication principle, the total number of seating arrangements is 720×6=4320720 \times 6 = 4320.

Adım Adım Çözüm

1
Group the restricted items into a single block
The 33 girls form 11 unit. Combined with the 55 boys, there are 5+1=65 + 1 = 6 units to arrange.
Since all 33 girls must sit together, treating them as a single block ensures they are not separated.
2
Calculate the linear arrangements of the combined units
The 66 units can be arranged in 6!=7206! = 720 ways.
The number of distinct ways to arrange nn items in a line is n!n!.
3
Calculate internal arrangements of the girls' block
The 33 girls can be arranged among themselves in 3!=63! = 6 ways.
The 33 girls inside the block are distinct individuals and can swap positions.
4
Apply the fundamental counting principle
Total arrangements = 6!×3!=720×6=43206! \times 3! = 720 \times 6 = 4320.
The total number of arrangements is the product of external block arrangements and internal block arrangements.

Anahtar Kavram

Permutations with grouping constraints (string method)
Soru 6Soru →

Calculate the total number of distinct four-digit numbers that can be formed using the digits 1,2,3,4,5,6,1, 2, 3, 4, 5, 6, and 77 without repetition, such that the resulting number is divisible by either 44 or 55.

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Cevap: 320

Cevap

The total number of such four-digit numbers is 320320.
The total number of valid four-digit numbers is found by adding the count of numbers divisible by 55 (120120) to the count of numbers divisible by 44 (200200). Since a number ending in 55 is odd, it cannot be divisible by 44, making the two conditions mutually exclusive. Thus, the total count is 120+200=320120 + 200 = 320.

Adım Adım Çözüm

1
Calculate the number of four-digit numbers divisible by 55.
For a number to be divisible by 55, its units digit must be 55 (since 00 is not available). There is 11 choice for the units digit. The remaining 33 positions are filled from the remaining 66 available digits in 6P3=6×5×4=120^{6}P_{3} = 6 \times 5 \times 4 = 120 ways.
Divisibility by 55 requires the last digit to be 55.
2
Calculate the number of four-digit numbers divisible by 44.
A number is divisible by 44 if its last two digits form a multiple of 44. Using distinct digits from {1,2,3,4,5,6,7}\{1, 2, 3, 4, 5, 6, 7\}, the valid two-digit endings are 12,16,24,32,36,52,56,64,72,12, 16, 24, 32, 36, 52, 56, 64, 72, and 7676 (1010 valid pairs). For each pair, the first two positions are filled from the remaining 55 digits in 5P2=5×4=20^{5}P_{2} = 5 \times 4 = 20 ways. Thus, total ways = 10×20=20010 \times 20 = 200.
Divisibility by 44 depends entirely on the last two digits.
3
Check for overlap (numbers divisible by both 44 and 55).
A number divisible by 55 must end in 55, which is an odd digit. All multiples of 44 must end in an even digit. Hence, no number is divisible by both 44 and 55 in this set. The overlap is 00.
The two events are mutually exclusive.
4
Apply the addition principle of counting.
Total = 120+200−0=320120 + 200 - 0 = 320.
Add the counts of the two mutually exclusive sets.

Anahtar Kavram

Restricted Permutations and Mutually Exclusive Events
Tahmini Süre:1m 30s
Soru 7Soru →

Find the number of distinct arrangements of the letters of the word PARALLEL\text{PARALLEL} such that no two letters ’L’\text{'L'} are adjacent.

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Cevap: 1200

Cevap

1200
To ensure no two letters 'L' are adjacent, we use the gap method. First, arrange the 5 non-L letters (P, A, R, A, E). Because 'A' repeats twice, there are 5! / 2! = 60 distinct arrangements. These 5 letters form 6 available gaps (including the ends). Selecting 3 of these 6 gaps to insert the 3 identical 'L's can be done in C(6, 3) = 20 ways. Multiplying these gives 60 × 20 = 1200 valid arrangements.

Adım Adım Çözüm

1
Count the frequency of each letter in the word PARALLEL.
The word has 8 letters: 1 P, 2 A's, 1 R, 3 L's, and 1 E.
Recognizing repeated elements is essential for permutations with duplicates.
2
Arrange the non-restricted letters (P, A, R, A, E).
Number of arrangements = 5! / 2! = 60.
The letter 'A' is repeated twice, so we divide 5! by 2!.
3
Calculate the number of available gaps for placing the 3 'L's so that no two are adjacent.
5 arranged letters create 6 gap positions. Choosing 3 gaps gives C(6, 3) = (6 × 5 × 4) / (3 × 2 × 1) = 20 ways.
Placing at most one 'L' per gap guarantees that no two 'L's will be adjacent.
4
Multiply the number of arrangements of non-L letters by the gap choices.
Total arrangements = 60 × 20 = 1200.
By the fundamental counting principle, total arrangements equal the product of independent choices.

Anahtar Kavram

Permutations with repeated elements and non-adjacency constraints using the Gap Method
Tahmini Süre:1m 30s
Soru 8Soru →

How many distinct five-digit even numbers greater than 50,00050,000 can be formed using the digits 2,3,5,6,7,2, 3, 5, 6, 7, and 88 if no digit may be repeated in any number?

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Cevap: 240

Cevap

The total number of distinct five-digit even numbers greater than 50,000 is 240.
The correct total is 240. Because the first digit requirement (greater than 50,000 requires 5, 6, 7, or 8) and the last digit requirement (even requires 2, 6, or 8) share common elements (6 and 8), the problem must be split into two mutually exclusive cases: when the first digit is odd (2 choices for d1d_1, 3 for d5d_5, and 4P3=24^4P_3 = 24 for the middle digits, giving 2×3×24=1442 \times 3 \times 24 = 144) and when the first digit is even (2 choices for d1d_1, 2 for d5d_5, and 4P3=24^4P_3 = 24 for the middle digits, giving 2×2×24=962 \times 2 \times 24 = 96). Adding these yields 144+96=240144 + 96 = 240.

Adım Adım Çözüm

1
Identify constraints on the positions of a five-digit number d1d2d3d4d5d_1 d_2 d_3 d_4 d_5.
For the number to be greater than 50,000, d1∈{5,6,7,8}d_1 \in \{5, 6, 7, 8\}. For the number to be even, d5∈{2,6,8}d_5 \in \{2, 6, 8\}.
The first digit determines the magnitude and the last digit determines parity.
2
Analyze Case 1 where the first digit d1d_1 is odd (55 or 77).
There are 2 choices for d1d_1. All 3 even digits remain available for d5d_5 (3 choices). The remaining 3 middle positions (d2,d3,d4d_2, d_3, d_4) can be arranged from the 4 remaining unused digits in 4P3=24^4P_3 = 24 ways. Total for Case 1 = 2×3×24=1442 \times 3 \times 24 = 144.
Choosing an odd first digit does not reduce the pool of available even digits for the units position.
3
Analyze Case 2 where the first digit d1d_1 is even (66 or 88).
There are 2 choices for d1d_1. Since one even digit is used in d1d_1, 2 even digits remain for d5d_5 (2 choices). The 3 middle positions are filled from the 4 remaining digits in 4P3=24^4P_3 = 24 ways. Total for Case 2 = 2×2×24=962 \times 2 \times 24 = 96.
Selecting an even digit for the ten-thousands place directly reduces the available choices for the units place.
4
Sum the possibilities from both mutually exclusive cases.
Total valid numbers = 144+96=240144 + 96 = 240.
Applying the addition principle of counting over disjoint cases.

Anahtar Kavram

Conditional permutations with non-disjoint position restrictions
Soru 9Soru →

In a science competition, 33 distinct prizes (first, second, and third place) are to be awarded to 33 different students chosen from a group of 55 finalists. In how many different ways can these 33 prizes be awarded?

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Cevap: 60

Cevap

60 ways
Because the prizes are distinct, the order in which the finalists are selected matters. Calculating the number of arrangements of 33 students from 55 candidates is given by 5P3=5×4×3=60^{5}P_{3} = 5 \times 4 \times 3 = 60.

Adım Adım Çözüm

1
Determine if order matters
Since the prizes are distinct (1st, 2nd, and 3rd place), the order of assignment matters, making this a permutation problem.
Assigning distinct positions to individuals requires calculating permutations rather than combinations.
2
Apply the permutation formula nPr=n!(n−r)!^{n}P_{r} = \frac{n!}{(n-r)!}
5P3=5!(5−3)!=5!2!^{5}P_{3} = \frac{5!}{(5-3)!} = \frac{5!}{2!}
There are 55 total candidates (n=5n = 5) and 33 positions to fill (r=3r = 3).
3
Calculate the numeric value
5P3=5×4×3=60^{5}P_{3} = 5 \times 4 \times 3 = 60
Canceling 2!2! from the numerator and denominator leaves 5×4×3=605 \times 4 \times 3 = 60.

Anahtar Kavram

Linear permutation of r items selected from n distinct items
Tahmini Süre:45s
Soru 10Soru →

Four different Mathematics textbooks and three different Physics textbooks are to be arranged in a line on a shelf. In how many distinct ways can the books be arranged if all four Mathematics textbooks must be kept together?

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Cevap: 576

Cevap

576 distinct ways
Treating the four Mathematics textbooks as a single unit gives 4 items to arrange on the shelf (the Mathematics unit and the three individual Physics textbooks). These 4 items can be arranged in 4!=244! = 24 ways. Furthermore, the four distinct Mathematics textbooks within the unit can be arranged among themselves in 4!=244! = 24 ways. By the multiplication principle of counting, the total number of distinct arrangements is 24×24=57624 \times 24 = 576.

Adım Adım Çözüm

1
Group the Mathematics textbooks into a single block
1 Mathematics block + 3 individual Physics textbooks = 4 items to arrange.
Because all four Mathematics textbooks must remain together, they act as a single composite unit.
2
Calculate the arrangements of the 4 main items
4!=4×3×2×1=244! = 4 \times 3 \times 2 \times 1 = 24 ways.
There are 4 distinct items (the Mathematics block and 3 separate Physics books) to place in linear order.
3
Calculate internal permutations of the Mathematics textbooks inside their block
4!=4×3×2×1=244! = 4 \times 3 \times 2 \times 1 = 24 ways.
The 4 Mathematics textbooks can be arranged in different orders among themselves.
4
Apply the fundamental counting principle
Total arrangements = 24×24=57624 \times 24 = 576.
Multiply the number of block arrangements by the internal arrangements of the Mathematics textbooks.

Anahtar Kavram

Permutations with Restricted Grouping (Block Method)
Soru 11Soru →

In how many distinct ways can a president, a vice-president, and a secretary be chosen from a group of 77 candidates, assuming no candidate can hold more than one position?

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Cevap: 210

Cevap

The total number of distinct ways to choose the three officers from 7 candidates is 210.
Selecting 3 distinct officers from a group of 7 candidates requires ordering 3 individuals out of 7, which equals 7×6×5=2107 \times 6 \times 5 = 210 ways.

Adım Adım Çözüm

1
Identify whether the problem involves permutations or combinations
Order is important because the positions (President, Vice-President, Secretary) are distinct.
Selecting person X as President and person Y as Secretary is different from selecting person Y as President and person X as Secretary.
2
Calculate the permutation 7P3^7P_3
7P3=7×6×5=210^7P_3 = 7 \times 6 \times 5 = 210
There are 7 choices for President, 6 remaining choices for Vice-President, and 5 remaining choices for Secretary.

Anahtar Kavram

Permutations of n items taken r at a time
Tahmini Süre:45s
Soru 12Soru →

In how many distinct ways can the letters of the word KADUNA\text{KADUNA} be arranged?

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Cevap: 360360

Cevap

The total number of distinct arrangements is 360360.
The word KADUNA\text{KADUNA} consists of 66 total letters with the letter '\text{A}' repeated 22 times. Using the formula for permutations with identical items, the number of distinct arrangements is 6!2!=7202=360\frac{6!}{2!} = \frac{720}{2} = 360.

Adım Adım Çözüm

1
Count the total number of letters in the word KADUNA\text{KADUNA} and identify repetitions.
Total letters n=6n = 6. The letter '\text{A}' appears 22 times, while '\text{K}', '\text{D}', '\text{U}', and '\text{N}' each appear 11 time.
Arrangements of nn items with repeated elements require dividing n!n! by the factorial of the count of each repeated element.
2
Apply the permutation formula for repeated elements: P=n!p!P = \frac{n!}{p!} where pp is the frequency of the repeated letter.
P = \frac{6!}{2!} = \frac{720}{2} = 360
Dividing by 2!2! eliminates duplicate arrangements caused by swapping identical letters.

Anahtar Kavram

Permutations of items with repeated elements
Tahmini Süre:45s
Soru 13Soru →

A committee of 66 delegates is to be seated around a circular table for a conference. If two specific delegates must always sit next to each other, in how many distinct ways can the delegates be arranged?

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Cevap: 48

Cevap

48
The correct answer is 48 because treating the 2 restricted delegates as a single unit leaves 5 items to arrange around a circle, which yields (5−1)!=24(5-1)! = 24 arrangements. Since the 2 delegates can arrange themselves in 2!=22! = 2 ways within their block, the total number of arrangements is 24×2=4824 \times 2 = 48.

Adım Adım Çözüm

1
Group the restricted delegates into a single block
2 specified delegates are treated as 1 unit, leaving 4 remaining delegates, making a total of 5 items to arrange.
Since the two delegates must sit next to each other, treating them as a single entity ensures they remain together in all seating positions.
2
Calculate the circular arrangements of the 5 items
(5−1)!=4!=24(5 - 1)! = 4! = 24 ways.
The number of ways to arrange nn distinct items around a circular table is given by (n−1)!(n - 1)!.
3
Account for internal arrangements of the paired delegates
2!=22! = 2 ways.
The two delegates within the single block can swap positions between themselves.
4
Multiply the circular arrangements by the internal arrangements
24×2=4824 \times 2 = 48 total distinct arrangements.
By the fundamental counting principle, total arrangements equal the product of the independent steps.

Anahtar Kavram

Circular Permutations with Restrictions
Tahmini Süre:1m 15s
Soru 14Soru →

Given the permutation equation nP4=42⋅nP2{^{n}P_4} = 42 \cdot {^{n}P_2}, what is the positive integer value of nn?

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Cevap: 9

Cevap

The positive integer value of nn is 99.
Expanding nP4{^{n}P_4} as n(n−1)(n−2)(n−3)n(n-1)(n-2)(n-3) and nP2{^{n}P_2} as n(n−1)n(n-1) transforms the equation to n(n−1)(n−2)(n−3)=42n(n−1)n(n-1)(n-2)(n-3) = 42n(n-1). Dividing both sides by n(n−1)n(n-1) gives (n−2)(n−3)=42(n-2)(n-3) = 42, which simplifies to n2−5n−36=0n^2 - 5n - 36 = 0. Factoring this quadratic gives (n−9)(n+4)=0(n-9)(n+4) = 0. Since nn must be a positive integer greater than or equal to 44, the correct answer is 99.

Adım Adım Çözüm

1
Write out the expanded expressions for nP4{^{n}P_4} and nP2{^{n}P_2}.
nP4=n(n−1)(n−2)(n−3){^{n}P_4} = n(n-1)(n-2)(n-3) and nP2=n(n−1){^{n}P_2} = n(n-1).
By definition, nPr=n!(n−r)!=n(n−1)…(n−r+1){^{n}P_r} = \frac{n!}{(n-r)!} = n(n-1)\dots(n-r+1).
2
Set up the algebraic equation based on the given problem statement.
n(n−1)(n−2)(n−3)=42n(n−1)n(n-1)(n-2)(n-3) = 42n(n-1).
Substitute the expanded permutation formulas into the given identity.
3
Simplify the equation by dividing both sides by the non-zero common product n(n−1)n(n-1).
(n−2)(n−3)=42(n-2)(n-3) = 42.
Since n≥4n \ge 4, n(n−1)≠0n(n-1) \neq 0, so we can cancel these terms from both sides.
4
Expand and rearrange the resulting expression into a standard quadratic equation.
n2−5n−36=0n^2 - 5n - 36 = 0.
Expanding (n−2)(n−3)(n-2)(n-3) gives n2−5n+6n^2 - 5n + 6; subtracting 4242 yields n2−5n−36=0n^2 - 5n - 36 = 0.
5
Solve the quadratic equation for nn and discard non-physical roots.
n=9n = 9.
Factoring yields (n−9)(n+4)=0(n-9)(n+4) = 0, so n=9n = 9 or n=−4n = -4. A permutation requires nn to be a positive integer ≥4\ge 4, so n=9n = 9.

Anahtar Kavram

Solving algebraic equations involving permutations nPr=n!(n−r)!{^{n}P_r} = \frac{n!}{(n-r)!}
Soru 15Soru →

Find the value of rr such that 8Pr=6720{^{8}P_r} = 6720.

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Cevap: 5

Cevap

The value of rr is 5.
Expanding 8Pr{^{8}P_r} into consecutive decreasing factors starting from 8 yields 8×7×6×5×4=67208 \times 7 \times 6 \times 5 \times 4 = 6720. Counting the number of factors multiplied (8, 7, 6, 5, 4) gives exactly 5 factors, so r=5r = 5.

Adım Adım Çözüm

1
Write the formula for permutation 8Pr{^{8}P_r} as a product of descending integers.
8Pr=8×7×6×⋯×(8−r+1){^{8}P_r} = 8 \times 7 \times 6 \times \dots \times (8 - r + 1)
By definition, nPr{^{n}P_r} represents the product of rr consecutive factors starting from nn and decreasing by 1.
2
Perform sequential multiplication starting from 8 until reaching 6720.
Product of 5 factors: 8×7×6×5×4=67208 \times 7 \times 6 \times 5 \times 4 = 6720
Multiplying factors gives: 8×7=568 \times 7 = 56; 56×6=33656 \times 6 = 336; 336×5=1680336 \times 5 = 1680; 1680×4=67201680 \times 4 = 6720.
3
Count the number of terms in the product to find rr.
r=5r = 5
Since 5 consecutive integers were multiplied together to obtain 6720, the subset size rr is 5.

Anahtar Kavram

Permutations of nn distinct items taken rr at a time
Tahmini Süre:1m 0s