In how many ways 5 men and 4 women can be arranged in a line so that

Question

In how many ways 5 men and 4 women can be arranged in a line so that no two women are besides each other?

A. 120
B. 360
C. 600
D. 43,200
E. 46,900

Bunuel · Accepted Answer

Solution provided by sandeep800 is correct.

Consider this:
5 men can be arranged in 5! number of ways;

Consider one particular arrangement of these men and empty slots as follows: *m*m*m*m*m*. Now, if we place 4 women in either of 4 slots out of 6 then no two women will be together -->  C^4_6  is the # of ways to choose in which 4 slots out of 6 these women will be placed and 4! is # of arrangements of them in these slots.

So total # of arrangements is:  5!*C^4_6*4! , which is the same as  5!*P^4_6 .

As for 6p4:  P^4_6=\frac{6!}{(6-4)!}=360 . # of ways to choose k distinct object out of n distinct object when the order of the chosen objects matters is given by the permutation formula:  P^k_n=\frac{n!}{(n-k)!} .

Hope it's clear.

sandeep800 · Answer

first of all we will arrange 5 men in 5 ! ways
|m1|m2|m3|m4|m5| m represents men and | places where women can sit without being together

then we have 6 positions for 4 women which can be filled in 6p4 ways 
ans 6p4*5!
360*120
hope it works
thanx

Anonymous · Answer

Could you explain how 6p4 comes out to 360? I'm having some trouble with the math. Thanks!

vwjetty · Answer

true. i dont understand either.

sandeep800 · Answer

hey sorry Guys i dint mention formula.

in case u dint understand then its 6!/(6-4)! i.e 6!/2! so (6*5*4*3*2*1)/(2*1) so it comes out to be 360

.Bunuel has doen my part of work too..Thanx

Womiwom · Answer

I'm totally confused. This was supposed to be an easier probability question?

sandeep800 · Answer

just ask ur doubt where and in which part u r confused!!i will try my best to help u out!!
thanx

Shruti0805 · Answer

The question mentions 4 women need to be arranged in a line so that they aren't adjacent to each other. Why cant we arrange them as : M1 W1 M2 W2 M3 W3 M4 W4 M5 ? This was, I guess there would be 5! ways in which men can be arranged and 4! ways in which women can be arranged, so a total of 5!*4!, which isn't even an option

I see someone in the previous posts has mentioned that the no. of slots to be filled by the women is 6 :S Thoroughly confused.

matthewsmith_89 · Answer

Womiwom  this is not probability. This is combinations. Probability is the extent to which an event is likely to occur, measured by the ratio of the favorable cases to the whole number of cases possible. Combinations, on the other hand, is a selection of a given number of elements from a larger number without regard to their arrangement.

gurmukh · Answer

5 men can sit in a line in 5! Ways. Now there are six spaces on the side and between them. So first women has six choices to sit and second women 5 choices to sit and so and so forth.
Total number of choices =120× 6×5×4×3= 43200

Posted from my mobile device

sahylsharma · Answer

gurmukh  can you please explain the six spaces on the side and between them part ? How do we know that there are six spaces on the side and between them? Thanks.

gurmukh · Answer

Let five men are a,b,c,d,e

_a_b_c_d_e_

These are six spaces represented by dash
For first women there are six choice to sit and for second women there are five choices and so on and so forth...

Posted from my mobile device

kamrulhasan26043 · Answer

Hi,  sandeep800  could you please tell why we didn't consider this arrangement ( M   W   M   W   M   W   M   W   M ) rather used (W M W M W M W M W M W) this?

Did I miss anything inference here?

HouseStark · Answer

__ = REPRESENT SPACES WHERE WOMEN CAN BE PLACED TO SEPERATE 2 WOMEN

__M__M__M__M__M__   THIS 5 MEN CAN BE ARRANGE THEMSELVES IN  5!  WAYS

AS YOU HAVE NOTICED THERE ARE 6 UNOCCUPIED SPACES WHERE WOMEN CAN PLACE, BUT THERE ARE 6 PLACES AND WE HAVE ONLY 4 WOMEN TO ARRANGE

NO. OF WAYS TO CHOOSE THOSE 4 PLACES WHERE WOMEN CAN BE PLACED= 6C4=15  WAYS

NOW WE HAVE TO ARRANGE THOSE 4 WOMEN AMONG THEMSELVES= 4!

TOTAL NO. OF WAYS=5!*15*4!

Kartik2121 · Answer

The answer is 6C4*5!*4!

Kinshook · Answer

In how many ways 5 men and 4 women can be arranged in a line so that no two women are besides each other

Let us have the arrange as shown below: -

_M_M_M_M_M_

We can arrange 5 men and 4 women in 5!*4! ways and can chose 4 slots out of 6 in 6C4 ways

Total number of ways such that 5 men and 4 women can be arranged in a line so that no two women are besides each other = 5!*4!*6C4 = 5!4!6!/4!2! = 5!6!/2! = 43200

IMO D

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In how many ways 5 men and 4 women can be arranged in a line so that

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