prashants412 wrote:

A group of 4 boys and 3 girls is to be seated in a row. How many such arrangements are possible where no 2 girls sit together?

A.144

B.288

C.576

D.720

E.1440

Most logical and Correct approach would be as shown by Pushpit..

Where do you have restrictions- on 2 girls sitting together.

So first let the boys sit in chairs with a gap in between..

_b_b_b_b_

GIRLS- So there are 5 gaps in which to place three girls, so 5C3= \(\frac{5!}{3!2!}=10\) but 3 girls can be arranged in 3!ways..

So 10*3!=60

BOYS- Butfor each arrangement of these 60, 4 boys can interchange places and can be arranged in 4! Ways..

Total =60*4!=1440

But say the restrictions were on 2 boys sitting together..

Place the 3girl...

_g_g_g_

So for boys there are 4 seats and 4 of them and you can choose 4out of 4 in 4C4=1

Ans 3!*4!*1

How about _BB_B_B here also we place Girls in the dashes then no two girls will be together.

again the group of two boys can move around so we can have _B_BB_B or _B_B_BB.

Does the solution given up till now cover all these cases? Thank you.