Solve the problem.
-How many different three-number "combinations" are possible on a combination lock having 22 numbers on its dial? Assume that no numbers repeat. (Combination locks are really permutation
Locks.)

Question

Quizplus · Accepted Answer

The total number of different three-number "combinations" (permutations, since order matters) possible on a combination lock with 22 numbers, where no numbers repeat, is calculated using the permutation formula $P(n, r) = \frac{n!}{(n-r)!}$. Here, $n = 22$ (total numbers) and $r = 3$ (numbers to choose). Thus, $P(22, 3) = \frac{22!}{(22-3)!} = \frac{22!}{19!} = 22 \times 21 \times 20 = 9240$.

Solve the Problem 1.7556×1051.7556 \times 10 ^ { 5 }1.7556×105 Three-Number "Combinations" C)

Solve the Problem $1.7556 \times 10 ^ { 5 }$ Three-Number "Combinations"
C)