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M20-09

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M20-09 [#permalink]

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If a music class consists of 4 girls and 7 boys, in 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

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Official Solution:

If a music class consists of 4 girls and 7 boys, in 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

The answer is (total number of groups of 3) - (number of all-girl groups of 3) = \(C_{11}^{3} - C_{4}^{3} = \frac{11!}{3! 8!} - \frac{4!}{3! 1!} = 15*11 - 4 = 161\).

Answer: C
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Re: M20-09 [#permalink]

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New post 17 Sep 2016, 14:38
I tried to solve it a different way:

There are 3 spots in the group. 7 ways to make sure there is always a boy in the first spot (7 C 1). Now, rest of the 2 spots have to be filled. There are 10 people left now from 11 because we already reserved 1 spot for a boy. This would be 10 C 2.

Total number of ways = 7 * 10 C 2 = 7 * 45 = 315.

What am I missing here?

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Re: M20-09 [#permalink]

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New post 20 Sep 2016, 22:15
In your method, some of the combinations are counted multiple times.
Let's try to understand with a smaller problem.

If a music class consists of 2 girls and 2 boys, in how many ways can a group of 2 be formed if it has to include at least one boy?

Method1: \(\dbinom{4}{2}\) - \(\dbinom{2}{2}\) = 6 - 1 = 5 ways (correct method)

Method2: \(\dbinom{2}{1} \times \dbinom{3}{1}\) = 6 (Incorrect method)

Enumeration (Method2):

Two boys: B1, B2, and two girls: G1, G2

Pick B1 first, then possible combination for B1:
B1,G1
B1,G2
B1,B2

Now pick B2 , possible combination for B2:
B2,G1
B2,G2
B2,B1

B1,B2, and B2,B1 are the essentially same team.

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Re: M20-09 [#permalink]

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New post 20 Sep 2016, 23:56
Bunuel wrote:
If a music class consists of 4 girls and 7 boys, in 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


Hi,
Total 3 places are there, out of which one has to be a boy..
So one seat can be occupied by any of the 7 boys...
So my answer should be a multiple of 7..
Only C is possible..


Further solution..
Remaining 2 seats can be filled by 4 girls and (7-1) boys in three ways
Both boys or both girls or one each..
7(6C2/3 + 4C2 + 6C1/2 * 4C1) = 7(5 + 6 + 12) = 7*23= 161..
Division by 3 and 2 is to negate same sets formed..
C

Ofcourse best way is as done by Bunuel,
Total - possibility where no boy is there = 11C3 - 4C3 = 165 - 4 = 161
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Re: M20-09 [#permalink]

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New post 01 Jul 2017, 01:12
Hi,

Why cant i do it like this-
1. 1 Boy & 2 girls- (7C1)*(4C2)=42
2. 2 Boys & 1 Girl- (7C2)*(4C1)=84
3. 3 Boys & 0 Girl-(7C3)*(4C0)=35
Total No. of Ways for at least 1 Boy in the group of 3=42+84+35=161

Hope this also makes sense.

Thanks.
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Re: M20-09 [#permalink]

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New post 25 Sep 2017, 06:09
The answer must be option C. We need to find the number of ways in which at least one boy is chosen. The answer must be 11C3-4C3 = 161. Ie, the total number of ways in which three can be chosen from total minus the number of cases where in 3 are chosen from 4 girls(or no boy is chosen).
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Re: M20-09 [#permalink]

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New post 25 Sep 2017, 07:14
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(4C2 *7C2)+(4C1*7C2)+(4C0*7C3)=42+84+35=161
Hence D
Remember + is used when there are more ways of doing the same task in different ways.
and * is used when the task at hand is not complete .
Hope this helps

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Re: M20-09   [#permalink] 25 Sep 2017, 07:14
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