Bunuel wrote:

A few men and a few women are seated in a row. The number of men is one greater than the number of women. What is the total number of people?

(1) The people are seated so that no two women sit beside one another.

(2) The number of ways of arranging the seating positions of the men and women is 3(5!)(7!).

manishtank1988 wrote:

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

_________________

Mike McGarry

Magoosh Test Prep