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Problem 106

AMC 10/12, early questions

Combinatorics Difficulty 3.2 Find the answer

A school selects 4 teachers from 8 to teach in 4 remote areas at the same time (one person per area), where A and B cannot go together, and A and C can only go together or not go at all. The total number of different dispatch plans is \_\_\_\_\_\_ (answer in numbers).

Your answer

A number or a short expression. Spacing, $ signs and \frac vs / are all fine.

Official solution

The solution can be divided into two steps:

First step, select four teachers, which can be further divided into two categories:

- First category, if A goes, then C must go, and B must not go. There are $C_5^2 = 10$ different ways to choose.
- Second category, if A does not go, then C must not go, and B may or may not go. There are $C_6^4 = 15$ different ways to choose.

Therefore, there are $10 + 15 = 25$ different ways to choose the teachers.

Second step, the four selected teachers go to 4 remote areas, which can be arranged in $A_4^4 = 24$ ways.

Finally, multiplying the number of methods in the two steps, we get $25 \times 24 = 600$ .

Hence, the answer is $\boxed{600}$ .

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