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Let $n$ be the number of ways in which $5$ men and $7$ women can stand in a queue such that all the women stand consecutively. Let $m$ be the number of ways in which the same $12$ persons can stand in a queue such that exactly $6$ women stand consecutively. Then the value of $\frac{m}{n}$ is

  1. $5$
  2. $7$
  3. $\frac{5}{7}$
  4. $\frac{7}{5}$
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Case-1:

The number of ways 7 women can stand in queue consecutively is n = 6!*7!

Case-2:

First, select 6 out of 7 women. This can be done 7 ways.

Let the group of 6 women be G

Now we have G, 1 woman, 5 men

They can be arranged in 7!*6! ways, but there are some arrangements where 7 women stand consecutively.

To find that, Consider G and one woman as one unit, now the number of arrangements possible is 2!*6!*6!

Required number of arrangements where exactly 6 women stand consecutively is m = 7*(7!*6! - (2!*6!*6!)) = 7*5*6!*6! = 5*7!*6!

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6 places: M M M M  [W W W W W W W]   

n = 7! x 6! (7 women ways)

_ B _ B _ B _ B _ B _

[W W W W W W] (exactly 6 women stand consecutively) <-> [W]

m = 7C6 x (6C2 x 2!) x 6! x 5!

m/n = 5

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There are 12 slots to be filled in for the queue,now in the question in the first case, we are told that women require to stand up consecutively,so number of ways it can done is as follows

No of ways we choose 7 consecutive slots out of 12 slots= 6 ways

After that,we arrange the women in 7 ! ways,so men get arranged in 5 ! ways

That counts to 6x7!x 5! ( say it is n )

 

As per second case,  we can choose 6 slots out of 12 slots in 7 ways, having chosen the slots,now let us choose 6 women from 7 women in 7C6 ways,and make them arrange in 6! ways,so remaining person are arranged in 6 ! ways

Total no of ways= 7 x( 7C6 x 6!) x 6! ( say it is m)

 

We are now required find m/n which is 7

 

Option B is right

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