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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.

SUBHASH BISHNOI
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2 Answers2

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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?)

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For a counting argument that counts “ways” of doing something to answer a question of the form “how many things?”, it must be the case that each “thing” is produced by one and exactly one of the “ways.”

There are indeed $3640$ ways to execute a certain sequence of steps: choose one man, then choose $3$ other members. And every committee with at least one many will come up as one of the results. But each different “way” does not result in a different committee. In fact, you would be counting quite a few committees more than once among the $3640$ “ways of choosing.” For example, any specific committee with $3$ men would be created $3$ times among the $3640$ enumerated “ways,” once with each of the men as the “first chosen” one.

Steve Kass
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  • The first paragraph is already incorrect. If in a "counting method" I can count each "way" exactly three times, and I know it's exactly three times, that still works just fine. The issue here, as I explained, is that not all "ways" are counted the same number of times. –  Jul 31 '16 at 19:29
  • thank you guys it helped. So, basically I am counting the same committee multiple times. thank you @mathguy – SUBHASH BISHNOI Jul 31 '16 at 20:09
  • @mathguy I thought my phrasing would be more helpful to the O.P., because I believe his/her fundamental misunderstanding was to conflate “ways” with “things.” Of course, the result of a counting argument that counts “ways” that produce every “thing” exactly $k$ times each can be divided by $k$ to produce the answer to the “things” question. However, I think when students learn these techniques, the distinction between “ways” and “things” must be understood first. – Steve Kass Jul 31 '16 at 20:29
  • Agreed ... like 99%. I still do believe that accurate statements help even beginners. No worries! –  Jul 31 '16 at 20:34
  • Thank you both for explaining. – SUBHASH BISHNOI Aug 01 '16 at 08:54