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.