Permutations & Combinations — Honors Math practice questions

Question 1

Determine $n \in \mathbb{N}$, $n \ge 2$, such that $\binom{n}{2} = 15$.

Accepted Answer

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

Answer

$4$

Answer

$5$

Answer

$6$

Answer

$15$

Question 2

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

Accepted Answer

$5 \cdot 4 = 20$, which is $A_5^2$.

Answer

$10$

Answer

$20$

Answer

$25$

Answer

$32$

Question 3

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

Accepted Answer

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

Answer

$522$

Answer

$774$

Answer

$999$

Answer

$1220$

Question 4

The number of $3$-element subsets of a $5$-element set is:

Accepted Answer

$\binom{5}{3} = \dfrac{5!}{3! \cdot 2!} = 10$.

Answer

$5$

Answer

$10$

Answer

$15$

Answer

$60$

Question 5

The number of $3$-letter codes that can be formed using letters from $\{A, B, C, D, E\}$ without repetition is:

Accepted Answer

$A_5^3 = 5 \cdot 4 \cdot 3 = 60$.

Answer

$15$

Answer

$20$

Answer

$60$

Answer

$125$

Question 6

A menu offers $4$ appetizers, $5$ mains, and $3$ desserts. The number of distinct three-course meals is:

Accepted Answer

$4 \cdot 5 \cdot 3 = 60$.

Answer

$12$

Answer

$36$

Answer

$60$

Answer

$120$

Question 7

The number of ways to arrange $4$ distinct books on a shelf is:

Accepted Answer

$4! = 24$.

Answer

$4$

Answer

$16$

Answer

$24$

Answer

$256$

Question 8

The value of $5!$ is:

Accepted Answer

$5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120$.

Answer

$25$

Answer

$60$

Answer

$120$

Answer

$720$

Question 9

At a meeting, every pair of people shakes hands exactly once. If there are $45$ handshakes total, how many people are there?

Accepted Answer

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

Answer

$9$

Answer

$9.5$

Answer

$10$

Answer

$45$

Question 10

Using Pascal's identity $\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}$, compute $\binom{7}{3} + \binom{7}{4}$.

Accepted Answer

By Pascal: $\binom{7}{3} + \binom{7}{4} = \binom{8}{4} = \dfrac{8!}{4!\,4!} = 70$.

Answer

$35$

Answer

$56$

Answer

$70$

Answer

$128$

Permutations & Combinations practice questions — Honors Math

Problems & worked solutions

Problem 1 — Permutations & Combinations

Problem 2 — Permutations & Combinations

Problem 3 — Permutations & Combinations

Problem 4 — Permutations & Combinations

Problem 5 — Permutations & Combinations

Problem 6 — Permutations & Combinations

Problem 7 — Permutations & Combinations

Problem 8 — Permutations & Combinations

Problem 9 — Permutations & Combinations

Problem 10 — Permutations & Combinations