A few men and a few women are seated in a row. The number of men is on

Vyshak, VeritasPrepKarishma, Skywalker18, mikemcgarry, Abhishek009, msk0657, abhimahna, Bunuel, Engr2012
Can someone explain how this highlighted line be calculated?
Thanks

Dear manishtank1988,

I'm happy to respond.

My friend, here's how I would think through the problem.
Statement #1: a seating constraint, but no number specific. Insufficient by itself.
Statement #2: a number given, but the number of possible seating constraints (e.g women must sit in clusters of 2 or 3 only) is mind-boggling. Insufficient by itself.

Combined: we have a clear seating constraint, which will determine a specific number of seating arrangements for each number of people, and the former will increase rapidly for each one we add to the latter. Thus, specifying a particular number of seating arrangements determines the number of people. Sufficient together.

OA = (C)

This is GMAT DS, so on principle, I would refuse to do a single calculation in answering the question. Nevertheless, you would like to see the calculation.

Again, here's how I would think about it.

Let's say there are x women. Just for the visual, I am showing the x = 5 case. These are spaced apart:

_ _ _ F _ _ _ F _ _ _ F _ _ _ F _ _ _ F _ _ _

There must be at least one man in each of the (x - 1) spaces between them. Put (x - 1) men there.

_ _ _ F _ M _ F _ M _ F _ M _ F _ M _ F _ _ _

We have two men left. Where do we put these last two men? We have (x - 1) slots between the two women as well as two on the ends, for (x + 1) total slots. We could either put both of these two extra men together, or we could put them in two different slots.
If they are together, we could put them in (x + 1). Possible outcomes:

M M _ F _ M _ F _ M _ F _ M _ F _ M _ F _ _ _

_ _ _ F _ M _ F _ M M M _ F _ M _ F _ M _ F _ _ _

_ _ _ F _ M _ F _ M _ F _ M _ F _ M _ F _ M M

If we put the two extra men in two different slots, we would have (1/2)(x+1)(x) choices. Possible outcomes:

_ M _ F _ M _ F _ M _ F _ M M _ F _ M _ F _ _ _

_ _ _ F _ M M _ F _ M _ F _ M _ F _ M M _ F _ _ _

Total number of "men slots" = (x + 1) + (1/2)(x + 1)x

Multiply that by (x+ 1)!, and then multiply this by x! for the women.

I get a different equation. I get
total number of arrangements = [(x + 1) + (1/2)(x + 1)x]*(x+ 1)!*x!

If x = 5, then this becomes

total number of arrangements = (6 + (1/2)(6)(5))*6!*5! = 21*6!*5! = 3*7*6!*5! = 3*7!*5! = 3(5!)(7!)

Does all this make sense?
Mike

mikemcgarry
First of all, thanks a lot for responding. I appreciate your help.
But i still have a few questions -
[Q-1] I can understand (x+1) => 1st M of the remaining 2 Ms can take (x+1) positions and 2nd M can take [(x+1)-1] = x positions. I don't understand why are we getting this (1/2)?

If we put the two extra men in two different slots, we would have (1/2)(x+1)(x) choices. Possible outcomes:
_ M _ F _ M _ F _ M _ F _ M M _ F _ M _ F _ _ _
_ _ _ F _ M M _ F _ M _ F _ M _ F _ M M _ F _ _ _


[Q-2] Also, i am assuming that we are multiplying by
(x+1)! => because finally there are (x+1) seats for men and we can arrange them in (x+1)! ways &
x! => because finally there are x seats for women and we can arrange them in x! ways
Am i correct in assuming this?
Thanks and looking forward to your reply.

Dear manishtank1988,

I'm happy to respond.

For Q1,
We are going to put the two men in two different slots, and there are (x + 1) slots. It's perfectly true we have (x + 1) choices for the first slot, and x choices for the second slot.

Let's say that x = 10, so we have 11 choices for the first slot and 10 choices for the second slot. Think about these two particular results:
Result #1: first choice: seventh slot; second choice: second slot
Result #2: first choice: second slot; second choice: seventh slot
These two different choices result in the same outcome, because order of picking the slots doesn't matter--all that matters is the two resultant slots.

In fact, every single result is counted twice, once where the leftmost slot was counted before the rightmost slot, and vice versa. We divide by two to eliminate the duplicates. That's why there has to be a 1/2 in the formula. In this situation, the result would be (1/2)(11)(10) = 11*5 = 55.

In fact, think about the combinations formula.
\(nCr = \tfrac{n!}{r!(n-r)!}\)

For the r = 2 case, this becomes
\(nC2 = \tfrac{n!}{2(n-2)!}\)

For n = 11, this becomes
\(nC2 = \tfrac{11!}{2(9)!} = \tfrac{11*10*9!!}{2(9)!} = \tfrac{11*10}{2} = 11*5 = 55\)

Thus even the combinations formula gives us that 1/2.

Here are two blogs you may find helpful.
GMAT Permutations and Combinations
GMAT Math: Calculating Combinations

In fact, this is really handy shortcut to know: nC2 = n(n - 1)/2

For Q2, you are 100% correct.

Does all this make sense?
Mike

Can someone please explain why the answer is not B?

We know that the total number of people is 2w+1. Therefore, the number of ways to arrange the people is (2w+1)!

(2w+1)! = 3(5!)(7!), sufficient

The key here is that the question does not say that all the men are identical and all the women are identical. Why would anyone assume that?

hello Sir,

I am having problems conceptualising the arrangement.................. Can you please help me??

the way I see it, first we can place the women ...lets say 5.... that is 5!

Then we can place atleast 1 man between 2 women............So 4 out of 6 Men can be chosen in 6C4 ways and arranged in 4! ways...

So total arrangement is 5!*6C4*4!

How to arrange the last 2 men??

Its generic Ex .5 people and note it doesn't say Daniel, Mat and sandy were seated. So from this perspective, all people are identical

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