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I have this one question in statistics:

There are five women and six men in a group. From this group a committee of 4 is to be chosen. In how many ways can the committee be formed if the committee is to have at least 3 women in it?

I am not exactly sure on how to approach this question since I cannot find a way to determine how many object there will and how many times they will be taken so that I can use the combination formula. My worries is with the words at least 3 women, does this mean that there could be 4 women then?

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  • $\begingroup$ Is this a homework excise? Please add [self-study] tag and check stats.stackexchange.com/tags/self-study/info $\endgroup$
    – Tim
    Commented May 26, 2016 at 18:25
  • $\begingroup$ It's not evident what information you might be lacking: there are $5+6=11$ people and the committee is to have $4$ people. What "object" are you referring to, then, where you state you don't know how many there will [be]? And you are correct that in English "at least $x$" means "$x$ or more." $\endgroup$
    – whuber
    Commented May 26, 2016 at 18:28
  • $\begingroup$ I am given all the information. i am having trouble with using the formula. $\endgroup$
    – rert588
    Commented May 26, 2016 at 18:30
  • $\begingroup$ I calculated the answer:315. Is it correct. $\endgroup$
    – rert588
    Commented May 26, 2016 at 18:34
  • $\begingroup$ You might be having trouble because this question requires you to think about the situation and decompose the solution into a few parts. You don't just dump the numbers into a formula. Incidentally, since there are only $\binom{11}{4}=330$ possible committees, regardless of gender, $315$ looks quite a bit too large to be correct. $\endgroup$
    – whuber
    Commented May 26, 2016 at 18:34

2 Answers 2

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Consider two cases:
1. There are exactly three women on the committee:

c(5,3)*c(6,1)=60
  1. There are exactly four women on the committee:

    c(5,4)=5

Hence:

Solution = 60+5 = 65
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I did this:

Okay. I said:

How many ways committee can be formed without specification i.e No 3 women restriction e.t.c:

c(11, 4) = 330

After that I say: Number of combinations that 5 women that can be taken 3 at a time = c(5, 3) = 10

Then I say, number of combinations that 5 women can be taken 4 at a time = c(5, 4) = 5

Final answer = 330 - 10 - 5 = 315

Do people agree with this?

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    $\begingroup$ Do you really think your calculation makes sense? Think about it again. You might want to approach it the other way and determine how many committees there are with exactly 3 women, and how many there are with exactly 4 women. $\endgroup$ Commented May 26, 2016 at 19:02
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    $\begingroup$ No and no, but you only have 327 possible answers left (well, maybe 328 if you think 0 could be the answer). Think it through carefully, and show your logic. Think not only about the number of women on the committee, but also what the corresponding number of men would be. That is my last hint and correctness verdict rendering. $\endgroup$ Commented May 26, 2016 at 19:10
  • $\begingroup$ There can only be 1 or 0 men on the committee $\endgroup$
    – rert588
    Commented May 26, 2016 at 19:13

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