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In how many ways can a committee of 4 be selected from nine men so as to always include a particular man? I thought 9 nCr3. as we only calculate to choose for 3 men out of the nine total. How am I wrong? Don't we have the consider the total ways that the other 3 people can be combined with that particular person.

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    $\begingroup$ I think it would be more like 3 out of 8, as one is already fixed. $\endgroup$ – pinkwerther Oct 27 '14 at 20:21
  • $\begingroup$ in short, my concern is that this problem involves the fundamental counting principle. the probability of the 3 people are multiplied with that one person. $\endgroup$ – user159778 Oct 27 '14 at 20:23
  • $\begingroup$ i.e. how many ways can those 3 people be paired with that one particular person. would this need to be counted? $\endgroup$ – user159778 Oct 27 '14 at 20:24
  • $\begingroup$ The probability? Take a step back I'd recommend. The only question to answer is how many ways can you choose three additional people. Ordering is not important here. $\endgroup$ – pinkwerther Oct 27 '14 at 20:25
  • $\begingroup$ So Mr. X is fixed; pick one of the others: 8 possibilities; pick one of the remaining 7: 7 possibilities; pick one of the remaining 6: 6 possibilities. Which gives you $8\times 7\times 6$ possibilities over all ;) Sorry for spoiling :( You still need to get rid of the ordering here though $\endgroup$ – pinkwerther Oct 27 '14 at 20:26
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You are selecting the other 3 men from 8, not 9, possibilities.

So the answer is $\binom{8}{3} = 56$ not 84.

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