IB Maths AA

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Permutations and combinations practice questions — IB Maths AA

11 free multiple-choice problems on permutations and combinations, ordered to match IB Maths AA difficulty, each with a full worked solution. No account needed — and there's a fresh problem set every day.

Problems & worked solutions

Problem #0410 International

Problem 1Permutations and combinations

At a meeting, every pair of people shakes hands exactly once. If there are 4545 handshakes total, how many people are there?

Show answer & worked solution
  1. A. 99
  2. B. 9.59.5
  3. C. 1010✓ correct
  4. D. 4545

n(n1)2=45n(n1)=90n=10\dfrac{n(n-1)}{2} = 45 \Rightarrow n(n-1) = 90 \Rightarrow n = 10.

Problem #0409 International

Problem 2Permutations and combinations

Using Pascal's identity (nk)+(nk+1)=(n+1k+1)\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}, compute (73)+(74)\binom{7}{3} + \binom{7}{4}.

Show answer & worked solution
  1. A. 3535
  2. B. 5656
  3. C. 7070✓ correct
  4. D. 128128

By Pascal: (73)+(74)=(84)=8!4!4!=70\binom{7}{3} + \binom{7}{4} = \binom{8}{4} = \dfrac{8!}{4!\,4!} = 70.

Problem #0407 International

Problem 3Permutations and combinations

The number of two-digit numbers with distinct digits formable from {1,2,3,4,5}\{1, 2, 3, 4, 5\} is:

Show answer & worked solution
  1. A. 1010
  2. B. 2020✓ correct
  3. C. 2525
  4. D. 3232

54=205 \cdot 4 = 20, which is A52A_5^2.

Problem #0404 International

Problem 4Permutations and combinations

The number of 33-letter codes that can be formed using letters from {A,B,C,D,E}\{A, B, C, D, E\} without repetition is:

Show answer & worked solution
  1. A. 1515
  2. B. 2020
  3. C. 6060✓ correct
  4. D. 125125

A53=543=60A_5^3 = 5 \cdot 4 \cdot 3 = 60.

Problem #0405 International

Problem 5Permutations and combinations

The number of 33-element subsets of a 55-element set is:

Show answer & worked solution
  1. A. 55
  2. B. 1010✓ correct
  3. C. 1515
  4. D. 6060

(53)=5!3!2!=10\binom{5}{3} = \dfrac{5!}{3! \cdot 2!} = 10.

Problem #0406 International

Problem 6Permutations and combinations

Determine nNn \in \mathbb{N}, n2n \ge 2, such that (n2)=15\binom{n}{2} = 15.

Show answer & worked solution
  1. A. 44
  2. B. 55
  3. C. 66✓ correct
  4. D. 1515

n(n1)2=15n2n30=0n{6,5}\dfrac{n(n-1)}{2} = 15 \Rightarrow n^2 - n - 30 = 0 \Rightarrow n \in \{6, -5\}. Only n=6n = 6 is admissible.

Problem #0408 International

Problem 7Permutations and combinations

The sum of all distinct three-digit numbers that can be formed using the digits {2,2,5}\{2, 2, 5\} is:

Show answer & worked solution
  1. A. 522522
  2. B. 774774
  3. C. 999999✓ correct
  4. D. 12201220

The three distinct numbers are 225,252,522225, 252, 522. Sum =225+252+522=999= 225 + 252 + 522 = 999.

Problem #0402 International

Problem 8Permutations and combinations

The number of ways to arrange 44 distinct books on a shelf is:

Show answer & worked solution
  1. A. 44
  2. B. 1616
  3. C. 2424✓ correct
  4. D. 256256

4!=244! = 24.

Problem #0403 International

Problem 9Permutations and combinations

A menu offers 44 appetizers, 55 mains, and 33 desserts. The number of distinct three-course meals is:

Show answer & worked solution
  1. A. 1212
  2. B. 3636
  3. C. 6060✓ correct
  4. D. 120120

453=604 \cdot 5 \cdot 3 = 60.

Problem #0401 International

Problem 10Permutations and combinations

The value of 5!5! is:

Show answer & worked solution
  1. A. 2525
  2. B. 6060
  3. C. 120120✓ correct
  4. D. 720720

5!=54321=1205! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120.

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