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Approach is faulty: Experts please comment

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Approach is faulty: Experts please comment

by imskpwr » Fri Jul 21, 2017 7:22 am
Anthony and Michael sit on the six-member board of directors for company X. If the board is to be split up into 2 three-person subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony? "¨
A. 20% "¨B. 30% "¨C. 40% "¨D. 50% "¨E. 60%

My approach:
Michael and Anthony same:
1st group: M A _,
2nd group: _ _ _
In 1st group, Keeping Michael and Anthony same so next place we have 4 c1 and
in 2nd group, we have 3c3
so total ways is 4c1 * 3c3............................................(1)

Total cases:

1st group: _ _ _,
2nd group: _ _ _
In 1st group, we have 6c3 and
in 2nd group, we have 3c3
so total ways is 6c3 * 3c3............................................(2)


Therefore probability is (1) / (2)
ithat gives me 4c1/ 6c3 = 4/10 OR 40%

Is it the correct approach?
I think it is incorrect. Can some expert confirm
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by Ian Stewart » Fri Jul 21, 2017 7:43 am
Your approach is correct, though you could streamline it a bit by ignoring the second committee, which doesn't matter here. We have 5C2 ways of picking the other two people on Michael's committee, and when Anthony is on that committee, we have 4 ways to pick the third person, so the answer is 4/5C2 = 4/10.

Even faster: once you assign Michael to a committee, there are 5 positions left, 2 on Michael's committee, and 3 on the other committee. So if you now assign Anthony to a random position, there will be a 2/5 probability that position is on Michael's committee.
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by imskpwr » Fri Jul 21, 2017 8:09 am
Ian Stewart wrote:Your approach is correct, though you could streamline it a bit by ignoring the second committee, which doesn't matter here. We have 5C2 ways of picking the other two people on Michael's committee, and when Anthony is on that committee, we have 4 ways to pick the third person, so the answer is 4/5C2 = 4/10.

Even faster: once you assign Michael to a committee, there are 5 positions left, 2 on Michael's committee, and 3 on the other committee. So if you now assign Anthony to a random position, there will be a 2/5 probability that position is on Michael's committee.
But since in 6c3* 3c3, groups are equal, don't you agree that we have to divide by 2! to remove duplicate values.

Since, The number of ways in which (m + n) different things can be divided into two groups, one containing m items and the other containing n items is given by m+nCn = (m+n)!/ m!n!
In the above case, if m = n i.e. the groups are of same size then the total number of ways of dividing 2n distinct items into two equal groups is given by 2nCn/2!.This can be written as (2n)!/n!n!2!
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by Ian Stewart » Fri Jul 21, 2017 2:43 pm
If you're asked: "in how many ways can six people be divided into two committees of three, if the order of the two committees does not matter?" (which is not a kind of question I've ever seen on the GMAT, incidentally, though I have seen a couple of prep company questions like that) then you do need to divide by 2! = 2 at the end, because the order of the 2 committees is not supposed to be important. That's what you're doing above, when you use that formula. But if you do that when you calculate the denominator of the answer to the Michael/Anthony question, you similarly need to assume the order of the two subcommittees is unimportant when you work out the numerator, so in the numerator you'll also divide by 2! = 2, and those 2's will cancel out.

So your method is logically fine, just as long as you're consistent - if you want to proceed by assuming the order of the two committees doesn't matter, then it must not matter in every calculation you do.

It tends to be easier to solve probability questions (which is essentially what this problem is) if you assume order does matter, at least when that's possible, since that won't usually affect your answer. If, say, you have 7 red and 3 green marbles in a bag, and I ask: "if you stick your hands in the bag, and grab two marbles, and pull your hands out at the exact same time, what is the probability both marbles are red?" then by the wording of my question, I'm trying to suggest the order in which we select the two marbles doesn't matter. But if you put your two hands in a bag and grab two marbles, the probability both are red won't change if you take your hands out one at a time, or take both hands out simultaneously. So we can think about taking out one marble at a time, even if the question isn't telling us that's what we're doing. And that's essentially what I did in my solutions above - I assumed we had a first committee, which I decided was the committee containing Michael, and then worked out the probability Anthony was also on that committee.
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by Matt@VeritasPrep » Sun Jul 23, 2017 3:20 pm
Another idea would be treating Michael and Anthony as one person, then assigning the committees from there.

If Michael and Anthony are one person, then we have five people to assign to the committees. Our Michael/Anthony siamese twin can go to either of the two groups (2!), then we simply choose a person to fill out that group (4 choose 1), leaving us with 2! * (4 choose 1) arrangements.

From there, we simply find the ratio of Michael/Anthony Twin Arrangements : Total Arrangements, and we're done:

2 * (4 choose 1) / (6 choose 3) => 8/20 => 40%
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Re: Approach is faulty: Experts please comment

by Scott@TargetTestPrep » Fri Feb 21, 2020 7:30 am
imskpwr wrote: ↑
Fri Jul 21, 2017 7:22 am
 Anthony and Michael sit on the six-member board of directors for company X. If the board is to be split up into 2 three-person subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony? "¨
A. 20% "¨B. 30% "¨C. 40% "¨D. 50% "¨E. 60%

My approach:
Michael and Anthony same:
1st group: M A _,
2nd group: _ _ _
In 1st group, Keeping Michael and Anthony same so next place we have 4 c1 and
in 2nd group, we have 3c3
so total ways is 4c1 * 3c3............................................(1)

Total cases:

1st group: _ _ _,
2nd group: _ _ _
In 1st group, we have 6c3 and
in 2nd group, we have 3c3
so total ways is 6c3 * 3c3............................................(2)


Therefore probability is (1) / (2)
ithat gives me 4c1/ 6c3 = 4/10 OR 40%

Is it the correct approach?
I think it is incorrect. Can some expert confirm
Let’s first determine the number of ways 2 three-person subcommittees can be formed from 6 people. The number of ways 3 people can be selected from 6 people for the first committee is 6C3 = (6 x 5 x 4)/(3 x 2) = 20. The number of ways 3 people can be selected from the remaining 3 people for the second committee is 3C3 = 1. Thus, the number of ways 2 three-person subcommittees can be formed from 6 people is 20 x 1 = 20, if the order of selecting the committees matters. However, since the order of selecting the committees doesn’t matter, we have to divide by 2! = 2. Thus, the number of ways 2 three-person subcommittees can be formed from 6 people is 20/2 = 10.

Since only a total of 10 committees can be formed, we can list all of these committees and see how many of them have Anthony and Michael on the same committee. We can let A be Anthony, M be Michael, and B, C, D, and E be the other 4 people.

1) A-B-C, D-E-M
2) A-B-D, C-E-M
3) A-B-E, C-D-M
4) A-B-M, C-D-E
5) A-C-D, B-E-M
6) A-C-E, B-D-M
7) A-C-M, B-D-E
8) A-D-E, B-C-M
9) A-D-M, B-C-E
10) A-E-M, B-C-D

We can see that from the 10 committees that can be formed, 4 of them (in bold) include both Anthony and Michael. Thus, the probability is 4/10 = 40%

Answer: C

Scott Woodbury-Stewart
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