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I'm learning Permutations and Combinations and while trying to solve this simple question, I stuck on the solution:-

From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

The solution was to consider all three possibilities one by one- 3 men and 2 women, 4 men and 1 women, all 5 men. Calculate the combinations for each case and add them.

But I was trying it from a different approach- Firstly select 3 men from 7 (gives 35 combinations) and multiply it with the combinations of selecting 2 more committee members from 10 remaining members. So my answer came out to be 35 x 45 = 1575. But the answer is 756 which is not even a multiple of 35. So is my approach wrong? Why so?

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If you first select 3 men and then fill up with two arbitrary people, then you count each combination with exactly three men once, but others are counted repeatedly. For example, you count all-male committees 10 times, once for each way to choose the three first men in retrospect from the final five men.

Smaller numbers example: Given John, Jack and Jill, form a committee of two people with the constraint that the committee must have at least one man. Actually, there are three such committees possible (because the constraint is no constraint at all).
But your method would give four: First select John as male, then select either Jack or Jill as second. Or first select Jack as male, then select John or Jill as second. The committee "Jack and John" is counted twice by you.

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  • $\begingroup$ I don't understand your first paragraph, but your second paragraph, with the simpler example, was very helpful. Thanks. $\endgroup$ – a.real.human.being Oct 5 '14 at 1:00
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You’re counting every committee with $4$ men $\binom43=4$ times: if the men are A, B, C, and D, and the woman is F, you’re counting it once with A, B, and C as the first $3$ and D and F as the extra $2$, once with A, B, and D as the first $3$ and C and F as the extra $2$, once with A, C, and D as the first $3$ and B and F as the extra $2$, and once with B, C, and D as the first $3$ and A and F as the extra $2$.

Similarly, you’re counting every committee with $5$ men $\binom53=10$ times: if the men are A, B, C, D, and E, you’re counting the committee once with A, B, and C as the first $3$ and D and E as the extra $2$, and so one for every possible $3$-$2$ split of the five men.

Only the committees of $3$ men and $2$ women are counted correctly, exactly once each.

Since some committees are counted once, some four times, and some ten times, you don’t even get an answer that’s related to the correct one in some very simple way, as you would if, for example, you counted every committee twice. From the correct answer you know that there are $525$ committees with $3$ men, $210$ with $4$ men, and $21$ with $5$ men, and sure enough,

$$525+4\cdot210+10\cdot21=1575\;,$$

your answer.

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You interpretation seems to right unlike your calculation

as

$$\binom 73\cdot\binom 62+\binom 74\cdot\binom 61+\binom 75\cdot\binom 60=35\cdot15+35\cdot6+21\cdot1=756$$

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  • $\begingroup$ This was the actual solution given in the book. I tried it differently and wanted to know why was I wrong? There was a difference in approach to the solution. Thanks btw :-) $\endgroup$ – vish213 Jun 16 '13 at 9:09

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