# Simple combination question

Seven managers and eight sales reps volunteer to attend a seminar. How many ways could 5 people be selected to go if there must be at least one manager and one sales rep?

The correct answer is found by making up 4 cases:
1 manager, 4 reps
2 managers, 3 reps
3 managers, 2 reps
4 managers, 1 rep

or

$_7C_1 \times _8C_4 + _7C_2 \times _8C_3 + _7C_3 \times _8C_2 + _7C_4 \times _8C_1 = 2926$

My original guess was just to find ways to choose 1 manager from 8, 1 sales rep from 7 and the remaining 3 from the remaining 13 or $_8C_1 \times _7C_1 \times _{13}C_3 = 16016$
Why does this not work?

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Notice that by doing what you did you count combinations more than once. More figuratively: Let $M_1,M_2,\ldots,M_7$ be your seven managers, and $R_1,\ldots,R_8$ be the reps.

consider the combination $\langle M_1,M_2,R_1,R_2,R_3\rangle$. By using the method you used you encounter this combination at least twice (six times to be more exact, be I leave it to you to figure why): Once when you choose $M_1$ to be the one manager out of 8, and some $R_i$ as the rep, and you select $M_2$ as one of the 3 of 13 other employees, and again when $M_2$ is chosen to be the manager, and $M_1$ is selected as one of the 3 in 13.

This is why you got a lot more results than the correct answer

Hope this helps

Shai :)

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