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Case 2: Two boys, one girl (7 choose 2) * (4 choose 1) = 84

Case 3: One boy, two girls (7 choose 1) * (4 choose 2) = 42

Add them up 35 + 84 + 42 = 161 C

I don't really understand how Case 2 and 3 above are calculated. Any help please?

Thanks, Diana

The music class consists of 4 girls and 7 boys. How many ways can a group of 3 be formed if it has to include at least one boy? A. 155 B. 158 C. 161 D. 165 E. 172

Reverse Approach:

The # of groups with at least one boy equal to total groups of 3 that can be formed out of 11 people minus groups with all girls: \(C^3_{11}-C^3_4=161\).

Answer: C.

Direct Approach:

The # of groups with at least one boy equal to groups with one boy (and 2 girls) plus groups with 2 boys (and 1 girl) plus groups with 3 boys: \(C^1_{7}*C^2_4+C^2_{7}*C^1_4+C^3_{7}=161\).

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29 Mar 2015, 09:08

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Re: The music class consists of 4 girls and 7 boys. How many way [#permalink]

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31 Mar 2015, 11:47

bmwhype2 wrote:

The music class consists of 4 girls and 7 boys. How many ways can a group of 3 be formed if it has to include at least one boy?

A. 155 B. 158 C. 161 D. 165 E. 172

We want At least 1 boy = Total - all girls Now all 3 girls can be selected in 4C3 = 4 ways No. of ways in which 3 people can be selected out of 11 = !!C3 = 165 Required number of ways = 165 - 4 = 161.

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Re: The music class consists of 4 girls and 7 boys. How many way [#permalink]

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31 Mar 2015, 12:13

I often confuse distinct between n!/((n-k)!*k!) and simple n!/(n-k)!. In this case I spend 90 seconds using solving 3 components, got answer 420, find out that it s wrong and divided every component to k!. And I did it with a simple example (how many variants to take 2 from 4) That is not good, I suppose) Too much time, and big risk to make mistake. What can you advice?

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19 Jul 2016, 04:40

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