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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. $120$
asked in Quantitative Aptitude by (32 points) 1 3
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3 Answers

+1 vote
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
answered by (38 points)
+1 vote

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

answered by (450 points) 2 6
0
Nicely explained
0 votes
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.
answered by (18 points)
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