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

Question

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?

mahakmalik · Accepted Answer

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

EvaJager · Answer

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.

Answer A.

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.

BrentGMATPrepNow · Answer

Take the task of selecting the 3 committee members and  break it into stages.

Stage 1: Select the 3 twin pairs from which we will select 1 sibling each. 
There are 4 pairs of twins, and we must select 3 pairs. Since the order in which we select the 3 pairs does not matter, we can use COMBINATIONS
We can select 3 pairs from 4 pairs in 4C3 ways ( 4  ways)

Stage 2: Take one of the 3 selected pairs and choose 1 person to be on the committee. 
There are 2 people in the twin pair, so this stage can be accomplished in  2  ways.

Stage 3: Take another of the 3 selected pairs and choose 1 person to be on the committee. 
There are 2 people in the twin pair, so this stage can be accomplished in  2  ways.

Stage 4: Take the last of the 3 selected pairs and choose 1 person to be on the committee. 
There are 2 people in the twin pair, so this stage can be accomplished in  2  ways.

By the Fundamental Counting Principle (FCP) we can complete all 4 stages (and thus create a 3-person committee) in  (4)(2)(2)(2)  ways (= 32 ways)

Answer = A

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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Edvento · Answer

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.

voodoochild · Answer

Edvento - yes, I was referring to Fund. Counting Principle.... thanks...

lonemeteorz · Answer

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

manishkhare · Answer

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.

fskilnik · Answer

\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.

ScottTargetTestPrep · Answer

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.

Answer: A

CrackverbalGMAT · Answer

Number of committes with NO siblings = Total committees – Number of committees with one pair of siblings.

Total committes = Select any 3 persons out of 8 persons (4 pairs of twins means 8 persons) = 8C3 = 56. 
Answer option C can be eliminated.

Number of committes with one pair of sibling = Select one pair of twins of the 4 pairs AND select any one person from the remaining 6 persons = 4C1 * 6C1 = 4 * 6 = 24.
Answer option B can be eliminated.

Number of committees with NO siblings = 56 – 24 = 32.

The correct answer option is A.

Hope that helps!
Aravind B T

lavanya.18 · Answer

2^3*4C3= 2*2*2*4=32

GMAT1034 · Answer

ScottTargetTestPrep , could you help with the 3! dividing here? I couldnt understand why are we doing that. Because as for 6 and 4, we have already excluded the overlaps.

Bunuel · Answer

We divide by 3! because the counting method used (multiplying 8 * 6 * 4) considers the order in which members are chosen, but the committee itself is unordered (i.e., selecting A, B, and C is the same as selecting B, C, and A). Since there are 3! (or 6) ways to arrange any three selected members, we must divide by 3! to avoid overcounting.

napolean92728 · Answer

Step 1: Understanding the problem 
We have 4 pairs of twins. This means there are 8 individuals in total. A committee of 3 members needs to be formed, but no committee should include both members of any twin pair.
 Step 2: Choosing the individuals 
Since no committee can include both members of any twin pair, we can only choose one person from each pair. This means we need to select 3 individuals from the 4 pairs of twins, ensuring that no twin pair is chosen together.
 Step 3: Counting the choices 
 
  Selecting 3 pairs of twins : First, we choose which 3 pairs of twins will contribute members to the committee. The number of ways to choose 3 pairs from 4 pairs is given by the combination formula:
 \binom{4}{3} =4 
So, there are 4 ways to choose 3 pairs out of the 4 available pairs. 
  Choosing 1 member from each selected pair : For each of the 3 selected pairs, we need to choose 1 individual. Since each pair has 2 members, there are 2 choices for each pair. Therefore, for 3 selected pairs, the number of ways to choose one individual from each pair is:
 2×2×2=8    
 Step 4: Total number of ways 
To find the total number of ways to form the committee, we multiply the number of ways to select the pairs (4 ways) by the number of ways to choose 1 individual from each pair (8 ways):
 4×8=32 
Thus, the total number of ways to form the committee is  A. 32 .

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

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