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The music class consists of 4 girls and 7 boys. How many way

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The music class consists of 4 girls and 7 boys. How many way [#permalink]  16 Oct 2007, 10:50
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 30 Jun 2012, 03:04, edited 1 time in total.
Edited the question.
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Re: PS Combinatorics - Music Class [#permalink]  16 Oct 2007, 11:02
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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?

155
158
161
165
172

Classic combination problem
At least 1 boy = Total - all girls
All girls = C(4,3) = 4
Total combination = C(11,3) = 165
Ans = 165-4 = 161
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Case 1: All three boys
(7 choose 3) = 35

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
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Re: [#permalink]  29 Jun 2012, 12:47
jimjohn wrote:
Case 1: All three boys
(7 choose 3) = 35

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
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Re: Re: [#permalink]  30 Jun 2012, 03:03
Expert's post
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dianamao wrote:
jimjohn wrote:
Case 1: All three boys
(7 choose 3) = 35

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: $$C^3_{11}-C^3_4=161$$.

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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Re: The music class consists of 4 girls and 7 boys. How many way [#permalink]  30 Jun 2012, 03:50
Awesome. Thanks Bunuel!

Cheers,
Diana
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Concentration: Finance
GMAT 1: 710 Q48 V39
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Re: The music class consists of 4 girls and 7 boys. How many way [#permalink]  30 Dec 2013, 06:34
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

Reverse combinatorics approach is preferred method on this one
Note that at least 1 boy gives clue to using this method

All combinations - four girls = answer

All combinations is 11C3 = 165

All girls 4C3 = 4

So our answer is 165-4 = 161

Hope it helps
Cheers!

J
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Re: The music class consists of 4 girls and 7 boys. How many way [#permalink]  29 Mar 2015, 09:08
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Re: The music class consists of 4 girls and 7 boys. How many way [#permalink]  31 Mar 2015, 11:47
Expert's post
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]  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?
Re: The music class consists of 4 girls and 7 boys. How many way   [#permalink] 31 Mar 2015, 12:13
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