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?