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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


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

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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up vote 4 down vote accepted

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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Because you are counting same configuration multiple times in the second approach. Let M denote managers and R denote reps. Suppose you picked M1 and R1 first and M2, M3 and M4 from the rest. The same configuration is obtained by picking M2 and R1 first and M1, M3 and M4 from the rest.

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I understand. Then is my first approach the most organized, or efficient way to go about this (without using generating functions, of course) – Paul Belardi Apr 10 '11 at 18:26
This illistrates the difference between a combination and permutation. If managers are denoted A, B, ...,G and sales reps are denoted 1,2,...,8, your question is to find the number of different five digit combinations with at least one letter and one number (where the order of the five digits does NOT matter). This is what you computed in the first method (a combination problem). In your second method, you are counting the number of five digit combinations containing at least one letter and one number, where the order matters to some extent. This is a type of permutation problem. – user3180 Apr 10 '11 at 18:36
I think the most efficient way is to first count the number of choices without any constraint on managers and sales reps, then subtract the number with no managers and the number with no sales reps (there being no choices with no managers and no sales reps). – Robert Israel Apr 10 '11 at 20:32

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