Bill and Mitt sit on the twelve-member board of directors for a

Question

Bill and Mitt sit on the twelve-member board of directors for a certain company. If the board is to be split up into four three-person subcommittees, what percent of all the possible subcommittees that exclude Bill also exclude Mitt?

A. \(43 \frac{2}{11} \)%

B. \(54 \)%

C. \(63 \frac{7}{11}\)%

D. \(72\frac{ 8}{11}\)%

E. \(75\)%

A.  43 \frac{2}{11}  %

B.  54  %

C.  63 \frac{7}{11} %

D.  72\frac{ 8}{11} %

E.  75 %

CrackverbalGMAT · Answer

Total number of committees that can be formed:

Choose 4 for the first committee out of 12, then choose another 4 out of the remaining 8 and then the last 4 will go into the last committee = 12C4 * 8C4 * 4C4 = 495 * 70 * 1 = 34650

Let Bill and Mitt be already there in the 1st committee. Then 2 more can be chosen out of 10 in 10C2 ways. Then choose 4 for the next committee out of 8 in 8C4 ways and the last 4 go into the last committee in 4C4 ways.
 = 10C2 * 8C4 * 4C4 = 3150

Similarly Bill and mitt can be chosen in the 2nd sub committee in 3150 ways (10C4 * 6C2 * 4C4) and in another 3150 ways when chosen into the last committee (10C4 * 6C4 * 4C4).

Total combinations where Bill and Mitt are always together = 3150 * 3 = 9150

Total Number of ways in which Bill and Mitt are never together = 34650 - 9150 = 2500

% =  \frac{25200}{34650} * 100 = \frac{800}{11} = 72\frac{8}{11}

Option D

Arun Kumar

stne · Answer

CrackVerbalGMAT , 
Thank you for the solution but Small error here!
Question says four 3 person committee i.e. --- --- --- --- ( There are four committees each containing 3 persons )
But question has been done using three 4 person committee i.e. ---- ---- ---- ( There are 3 committees each containing 4 person)

Also in the last portion 34650-9150=25500 but you have taken 25200 and arrived at the answer.

Can you please revise the solution.

Thank you.

Regor60 · Answer

Since Bill has an equal chance of being in any of the 4 committees, he is in 1/4 of them and NOT in:

3/4

Because there are two spots on a committee that includes Bill, Mitt has a 2/11 chance of being included with Bill, meaning he has a:

9/11 chance of being among the 3 committees that exclude Bill

Mitt himself can only be on 1 of these three committees so he is excluded from 2/3 of them

So overall:

9/11 * 2/3 = 18/33 = 24/44 of the committees will exclude both Bill and Mitt.

So, the percentage of committees that exclude Bill which also exclude Mitt is:

(24/44)/(3/4) = (4/3)*(24/44)

= 8/11 = 72 8/11 percent

Posted from my mobile device

mcrea · Answer

My way:

Combinations in which a subcommittee can be formed with neither Mitt nor Bill over combinations in which a subcommittee can be formed only without Bill. Thus, possible combinations of 3 people out of a group of 10 people (12 minus Mit and Bill) over possible combinations of group of 3 people out of a group of 11 people (12 minus Bill).

(10C3 / 11C3) * 100 = (120 / 165) * 100 = 72 8/11

IMO D

Bill and Mitt sit on the twelve-member board of directors for a

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