Question

There are three girls in a class of $10$ students. The numbers of different ways in which they can be seated in a row such that no two of the three girls are together is

• A. $7!{\times }^6{p}_3$
• B. $7!{\times }^8{p}_3$
• C. $7!\times 3!$
• D. $\frac{10!}{3!7!}$

Answer 1

Answer: (B)

$7!{\times }^8{p}_3$

_B_B_B_B_B_B_B

B$=$boy

$=$Possible places for girls

No. of permutations of boys$=7!$

girl can permute in $11$ places i.e., ${}^8{p}_3$

keep unoccupied seats vacant

Ans. $7!{\times }^8{p}_3$.