Track / Stage 3 / 42 of 260 #42 of 2000

Problem 42

AMC 10/12, early questions

Combinatorics Difficulty 3.1 Find the answer

Three volunteers, A, B, and C, are scheduled to participate in a volunteer activity from Monday to Friday, with each person participating for one day and at most one person scheduled per day. It is required that A is scheduled before the other two. How many different scheduling methods are there? (Answer with a number)

Your answer

Official solution

To solve this problem, we first consider the position of A. Since A must be scheduled before B and C, and each person participates on a different day, there are three scenarios for A's participation:

1. If A is scheduled on Monday, then there are $4$ days left for B and C, which means there are $4 \times 3 = 12$ ways to schedule B and C.
2. If A is scheduled on Tuesday, then there are $3$ days left for B and C, which means there are $3 \times 2 = 6$ ways to schedule B and C.
3. If A is scheduled on Wednesday, then there are $2$ days left for B and C, which means there is only $2 \times 1 = 2$ way to schedule B and C.

Adding these scenarios together, the total number of different scheduling methods is $12 + 6 + 2 = 20$ .

Therefore, the answer is $\boxed{20}$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.