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How many ways $4$ men and $3$ women can speak at a conference so that women speakers are grouped together?

1. $720$

2. $488$

3. $360$

4. $120$
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let men speakers be M and women speakers be W then,

 M M M M W W W

Now this problem is boil down to 5 Box problem where in one Box there are 3 W So,

For arranging 3 W  there are 3! ways.

For arranging 5 Boxes there are 5! ways so, Total no. of ways = 5! x 3! =720

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Assume all 3 women as a single group.

Now, you can arrange 4 men and 1 group of women in a particular order in 5! Ways.

T e internal arrangement of women can be done in 3! Ways.

Total possible arrangements = 5! * 3! = 120 * 6 = 720
OPTION-->A

Since the order of seating matters ,we are going to choose permutations here.

Suppose ABCDE are 4 mens and one women respectively, ABCDE =! ABCED.

So coming to the question,4 men and 3 women together.

Consider 3 women as a single block and total no of persons =5.

For first position we have 5 choices ,next 4..3..2..1.

We can arrange in 5! ways.Since the block of women is jumbled across,the 3 women can be placed themselves along in 3! ways.

So the final answer will be 3!.5!=6*120=720.
assume you have tied all the women with a rope in order to ensure that they always remain together. Now , we have 5 groups i.e. 4 men and one group of women consisting of 3 women which is tied together.

We can arrange group of 5 people in 5! ways

Now, in women group which are tied ,among them 3 different women are there which can be arranged in 3! ways

Now, both of them have to be together = 5! * 3! = 720
answered by (74 points) 1 3