From a group of 8 men and 8 women, 4 persons are to be selected to form a committee so that at least 1 man is there on the committee. In how many ways can it be done? ( this is a repeat question ).
I found 2 solution to this:
1. combination of (1 male+ 3 female)+C(2M+2F)+(3M+1F)+(4M+0F) = 1750
2. Combination of all possible selection – combination of all Female i.e. 16C4-8C4 = 1750.
My question is can why cannot we write the solution as:
8C1*15C3 = 3640.
i.e selection of 1 men from 8 and then selection of rest 3 from 15 in any order.
as the first part will make sure that we have at least 1 man in our committee.
Please explain if i am missing anything as it doesn't match with other answers.
Sorry for the repeat question and formatting.
Best Answer
Because in your "proposed solution" you are counting some of the possible committees more than once. Worse, they are not all counted the same number of times. A committee with exactly one man is not double-counted. A committee with two men is counted twice in your total of 3640. One with three men is counted three times, and one with four men is counted four times. (A committee is counted, in your approach, as many times as there are men on the committee - WHICH of the two, three or four men on the committee is the one you singled out?)