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# If there are 4 pairs of twins, and a committee will be

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If there are 4 pairs of twins, and a committee will be  [#permalink]

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Updated on: 29 Jul 2012, 14:54
4
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45% (medium)

Question Stats:

60% (01:38) correct 40% (02:02) wrong based on 273 sessions

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If there are 4 pairs of twins, and a committee will be formed with 3 members. In how many ways this committee formed in a way that no siblings in a group?

A. 32
B. 24
C. 56
D. 44
E. 40

The problem can be solved using Combinations. I want to understand whether the above problem can be solved using FCP? If not, then why? [/b] Please don't post solutions using Combinations. There are a million other threads on the same topic. Thanks

Originally posted by voodoochild on 29 Jul 2012, 09:06.
Last edited by Bunuel on 29 Jul 2012, 14:54, edited 2 times in total.
Edited the question and renamed the topic.
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Re: If there are 4 pairs of twins, and a committee will be  [#permalink]

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30 Jul 2012, 02:16
3
voodoochild wrote:
If there are 4 pairs of twins, and a committee will be formed with 3 members. In how many ways this committee formed in a way that no siblings in a group?

A. 32
B. 24
C. 56
D. 44
E. 40

The problem can be solved using Combinations. I want to understand whether the above problem can be solved using FCP? If not, then why? [/b] Please don't post solutions using Combinations. There are a million other threads on the same topic. Thanks

What do you mean by FCP? Factorial, combinations, permutations?
But these are typical questions for the use of such mathematical tools.
Anyway, you have to use at least the multiplication principle, you cannot do without it.

Here is an approach:
The first member can be anybody, so 8 possibilities.
The second member cannot be the sibling of the previous member, therefore 6 possibilities (anybody from the remaining 3 pairs).
Finally, the third member, can be chosen from the remaining 2 pairs, so 4 possibilities.
This would give us 8*6*4 possibilities, but in this case we don't care about the order in which we choose them, we have to divide the product we obtained by 6(=3!), which is the number of possibilities we can chose the same three distinct members, say ABC (it can be BAC, CAB,...). You can count those possibilities for such a small number as 3, but why avoid permutations? Is it better to start listing the possibilities instead of just accepting the already proven result?

So, the correct answer is 8*4=32.

Avoiding FCP, you can start listing all the possibilities, solve just by brute force. FCP were developed exactly to deal with counting problems, to avoid lengthy counting. There are no other methods in mathematics for these type of questions.
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Re: If there are 4 pairs of twins, and a committee will be  [#permalink]

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30 Jul 2012, 14:45
Hi EvaJagger,

I think by FCP, voodoochild meant Fundamental Counting Principle... which is essentially how Permutations and Combinations formulas are derived from, which is where the concept of factorial had evolved. And on that front, I think you have explained pretty well. 8*6*4/6 it is.
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Re: If there are 4 pairs of twins, and a committee will be  [#permalink]

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31 Jul 2012, 06:05
Edvento - yes, I was referring to Fund. Counting Principle.... thanks...
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Re: If there are 4 pairs of twins, and a committee will be  [#permalink]

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25 Sep 2015, 19:23
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3
Firstly, there are 4 ways to choose 3 groups from the 4 groups. Hence 4C3.

Next, with the 3 groups chosen, there are 2^3 choices of choosing either one of the siblings.

Hence, the total number of ways are 4C3 * (2^3) = 32
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Re: If there are 4 pairs of twins, and a committee will be  [#permalink]

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01 Oct 2015, 19:28
Could you post a combinatorics solution?
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If there are 4 pairs of twins, and a committee will be  [#permalink]

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02 Oct 2015, 03:58
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total possibilities including twins are$$8c3$$ = 56

first choose any pair of twins i.e $$4c1$$ then remaining position can be filled by of the 6 members.

total possibilities only twins= $$4c1$$ * $$6c1$$=24

no twins included = 56-24=32

+1 Kudo if you like the solution
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Re: If there are 4 pairs of twins, and a committee will be  [#permalink]

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05 Oct 2015, 21:33
1
voodoochild,
Number of Members =8
First member can be selected in 8 ways
Second member can be selected in 6 ways (Siblings cannot be selected)
Third member can be selected in 4 ways (Siblings cannot be selected)

Number of ways =8X6X4=192

Observe that which sibling is selected first is immaterial i.e. ABC=BCA=CAB
Any one can be first or last .We need to remove the duplicate selections .

Number of Duplicate selections =6( 3X2X1 .We need to select 3 members)

Hence total number of ways =192/6 =32
Option A.
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Re: If there are 4 pairs of twins, and a committee will be  [#permalink]

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23 Dec 2015, 04:12
voodoochild wrote:
If there are 4 pairs of twins, and a committee will be formed with 3 members. In how many ways this committee formed in a way that no siblings in a group?

A. 32
B. 24
C. 56
D. 44
E. 40

The problem can be solved using Combinations. I want to understand whether the above problem can be solved using FCP? If not, then why? [/b] Please don't post solutions using Combinations. There are a million other threads on the same topic. Thanks

1C2 x 1C2 x 1C2 x 3C4 = 32 A

no siblings mean, 1 guy from the 1st pair , 1 guy from 2nd pair, 1 guy from 3rd pair

we choose from 3 pairs but got 4 pairs
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Re: If there are 4 pairs of twins, and a committee will be  [#permalink]

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20 Jan 2017, 08:42
voodoochild wrote:
If there are 4 pairs of twins, and a committee will be formed with 3 members. In how many ways this committee formed in a way that no siblings in a group?

A. 32
B. 24
C. 56
D. 44
E. 40

The problem can be solved using Combinations. I want to understand whether the above problem can be solved using FCP? If not, then why? [/b] Please don't post solutions using Combinations. There are a million other threads on the same topic. Thanks

Hi,

Consider twins as one unit so 3 members committee will be formed by using 4 units is 4C3 =4

Now consider there are 4 pairs no siblings in a group? i,e 2*2*2=8

Bunuel is this right ?
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Re: If there are 4 pairs of twins, and a committee will be  [#permalink]

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20 Jan 2017, 09:09
mahakmalik wrote:
total possibilities including twins are$$8c3$$ = 56

first choose any pair of twins i.e $$4c1$$ then remaining position can be filled by of the 6 members.

total possibilities only twins= $$4c1$$ * $$6c1$$=24

no twins included = 56-24=32

+1 Kudo if you like the solution

Hello, can you explain me why the total possibilities for only twins is 4c1 * 6C1 i can understand why 4c1. Thank you in advance.
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Re: If there are 4 pairs of twins, and a committee will be  [#permalink]

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03 Dec 2018, 13:28
2
voodoochild wrote:
If there are 4 pairs of twins, and a committee will be formed with 3 members. In how many ways this committee formed in a way that no siblings in a group?

A. 32
B. 24
C. 56
D. 44
E. 40

$$\left. \matrix{ \# \,\,{\rm{total}}\,\,{\rm{committees}}\,\,{\rm{ = }}\,\,\,{\rm{C}}\left( {8,3} \right) = {{8 \cdot 7 \cdot 6} \over {3!}} = 56 \hfill \cr \# \,\,{\rm{committees}}\,\,{\rm{with}}\,\,{\rm{siblings}}\,\,{\rm{ = }}\,\,\,\underbrace {\,\,\,4\,\,\,}_{{\rm{choice}}\,\,{\rm{of}}\,\,{\rm{pair}}\,\,{\rm{of}}\,\,{\rm{twins}}} \cdot \underbrace {\,\,\,6\,\,}_{{\rm{choice}}\,\,{\rm{out - of - the - pair}}} = 24\,\,\, \hfill \cr} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,?\,\,\, = \,\,\,56 - 24 = 32$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: If there are 4 pairs of twins, and a committee will be  [#permalink]

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22 Mar 2019, 09:31
voodoochild wrote:
If there are 4 pairs of twins, and a committee will be formed with 3 members. In how many ways this committee formed in a way that no siblings in a group?

A. 32
B. 24
C. 56
D. 44
E. 40

The number of ways to select the 3 pairs from 4 pairs is 4C3 = 4.

Since there can be no siblings on the board each twin can be selected in 2C1 ways, so:

2C1 x 2C1 x 2C1 = 2 x 2 x 2 = 8

So the total number of ways to select the committee is 4 x 8= 32.

Alternate Solution:

For the first member, there are 8 choices. Since the sibling of the first member cannot be chosen, there are 6 choices for the second member. By the same logic, there are 4 choices for the last member. Notice that the 8 x 6 x 4 choices count each committee 3! times; therefore, there are (8 x 6 x 4)/3! = 8 x 4 = 32 possible committees.

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Re: If there are 4 pairs of twins, and a committee will be   [#permalink] 22 Mar 2019, 09:31
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