I would be happy to explain.

Since the condition is that no 2 girls should sit together, we should ensure that there would be at least one boy between any 2 girls.

Now the 4 boys have already been placed in 4! ways.

After that we need to find the slots which can be filled by girls so that no 2 girls are together,

- B - B - B - B - If you observe carefully,

if the girls are made to sit on the 5 blanks seat represented by -, then no 2 girls would be together.So as we have 3 girls , we need to select 3 blanks seat out of 5 blanks as mentioned above, which can be done in 5C3 ways.

After the selection of 3 blanks seats, the girls can be arranged in 3! ways.

Hence the total number of ways would be 4!*5C3*3! = 1440.

I hope it is clear now, feel free to ask further in case of any doubts.

SalmanDard wrote:

GMATinsight wrote:

rishabhmishra wrote:

A group of four boys and three girls is to be seated in a row. how many such arrangements are possible where no girls sit together ?

A. 144

B. 288

C. 576

D. 720

E. 1440

Source:-

Experts global.

Between every two BOYS there must be A MAXIMUM OF one GIRL hence we the the arrangements will look as follows

- B - B - B - B - But since there are only 3 girls and 5 places available for them (shown by dashes above) so we need to select those 3 places for the girls out of 5

Selection of 3 places out of 5 for the girls = 5C3

Arrangement of the girls on 3 selected places = 3!

Arrangement of the boys on 4 places = 4!

Total Favourable outcomes = 5C3*3!*4! = 60*24 = 1440

Answer: option E

I am trying to understand why are you assuming 9 spaces to be filled? are we not supposed to fill 7 spaces (4 boys and 3 girls)? Kindly help me understand!

"B-B-B-B"

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