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navik [9.2K]
3 years ago
7

Best/correct answer gets brainliest!

Mathematics
2 answers:
Neporo4naja [7]3 years ago
3 0
I think the answer is 12.8 mm 
melisa1 [442]3 years ago
3 0
Answer for your question is 12.8
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How many possible committees of 4 people can be chosen from 15 men and 20 women so that at least two women should be on each com
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<span>If there has to be 2 men and 2 women, we know that we must take a group of 2 men out of the group of 15 men and a group of 2 women out of the group of 20 women. Therefore, we have:

(15 choose 2) x (20 choose 2)


(15 choose 2) = 105

(20 choose 2) = 190

190*105 = 19950

Therefore, there are 19950 ways to have a group of 4 with 2 men and 2women.</span>

 

<span>If there has to be 1 man and 3 women, we know that we must take a group of 1 man out of the group of 15 men and a group of 3 women out of the group of 20 women. Therefore, we have:

(15 choose 1) x (20 choose 3)


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(20 choose 3) = 1140

15*1140 = 17100

Therefore, there are 17100 ways to have a group of 4 with 3 women and 1 man.</span>

 

<span>We now find the total outcomes of having a group with 4 women.

We know this is the same as saying (20 choose 4) = 4845</span>

Therefore, there are 4845 ways to have a group of 4 with 4 women.

 

We now add the outcomes of 2 women, 3 women, and 4 women and get the total ways that a committee can have at least 2 women.

 

19950 + 17100 + 4845 = 41895 ways that there will be at least 2 women in the committee


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3 years ago
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