Consider m men in a row separated by empty places before and after each of them: *M*M*...*M*M*. So, we have m men and m + 1 empty places (stars). If we place women on that empty places then no two women will be together: any two of them will be separated by at least one men between them.

Choosing which places would women occupy = \(C^n_{m+1}=\frac{(m+1)!}{n!(m+1-n)!}\) (choosing n places out of m+1 empty slots).

Next, women themselves can be arranged on their places in n! ways, and men themselves can be arranged in m! ways. Thus, the final answer is \(\frac{(m+1)!}{n!(m+1-n)!}*n!*m!=\frac{(m+1)!m!}{(m+1-n)!}\).

Answer: B.
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Shouldn't empty spaces be m-1 otherwise the number of women being seated will be more than men (but m>n). Please clarify.

m + 1 empty places are just options for n women. Meaning that we are still choosing n destinations from m + 1 options.
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Suppose I'm taking m=3 and n=2
Then number of ways would be= Total arrangement - (arrangement in which women will be together)
which will be equal to 5! - 2*(4!) = 72

72 is higher than what I'm getting by putting the values in the formula.

What is wrong with this approach? or is there anything which I have missed?

m= no.of men
n= no.of women
Let just simply understand with an example as in question mentioned m>n
Lets assume m=10 & n=5

Let's place those men 1st in row

_A_B_C_D_E_F_G_H_I_J_

NO. OF WAYS TO ARRANGE THOSE 10 MEN=10!

YOU CAN SEE THOSE SPACES BETWEEN MEN.. IS THE PLACE YOU SUPPOSE TO PLACE THE WOMEN TO SEPARATE THEM FROM SITTING TOGETHER

No.of ways to select those 5 positions for women to select among 11 unoccupied spot=11C5

Now, time to arrange those 5 women=\(11C5*5!\)

Answer=\(10!*11C5*5!\)

=\(10!*\frac{11!}{5!(11-5)!}*5!\)

=\(10!*\frac{11!}{(11-5)!}\)

You can see it matches with the option B

\(m!*\frac{(m+1)!}{(m-n+1)!}\)

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Guys, I have a doubt about this question: why do I need to have one empty space BEFORE and one AFTER the men?
I mean, if I try to solve the question like this:

_M_M_M_ -> it WORKS, but if I try like M_M_M, it doesn't work.

I don't understand it, since in both cases I think I am getting what the prompt wants (m > n and no women together).
Could you please help me? Bunuel

Thank you all so much!

Solution:

We can use suitable numbers for m and n to figure out this problem. We can let m = 4 and n = 2 and let the four men be A, B, C, D and the two women be 1 and 2. If there are no restrictions, there are 6! = 720 arrangements since there are 6 people. Let’s now figure out the number of seating arrangements for the restriction (i.e., let’s suppose the two women must sit together). We can have:

ABCD12, ABC12D, AB12CD, A12BCD, and 12ABCD

However, for each the seating arrangements listed above, there are 4! x 2! = 24 x 2 = 48 arrangements since we can arrange the 4 men 4! ways and the 2 women 2! ways. Therefore, there are 5 x 48 = 240 arrangements if the two women must sit together. This means that there are 720 - 240 = 480 arrangements where no two women sit together. Now, let’s see which of the given choices give us the value of 480 when m = 4 and n = 2.

A) m!/[n!(m-n)!] = 4!/[2! x 2!] = 24/4 = 6 → This is not 480.

B) m!(m+1)!/(m-n+1)! = 4! x 5!/3! = 24 x 120/6 = 480 → This is 480. If none of the other choices yields the value of 72, then B must be the correct answer.

C) m!n!/[n!(m-n)!] = 4! x 2!/[2! x 2!] = 24 x 2/4 = 12 → This is not 480.

D) m!(m+1)!/[n!(m-n)!] = 4! x 5!/[2! x 2!] = 24 x 120/4 = 720 → This is not 480.

E) m!n!/(m-n+1)! = 4! x 2!/3! = 24 x 4/6 = 16 → This is not 480.

Answer: B
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