In a group of 3 girls and 4 boys, there are two boys B 1 and B 2 . The number of ways, in which these girls and boys can stand in a queue such that all the girls stand together, all the boys stand

Question

In a group of 3 girls and 4 boys, there are two boys $B_{1}$ and $B_{2}$ . The number of ways, in which these girls and boys can stand in a queue such that all the girls stand together, all the boys stand together, but $B_{1}$ and $B_{2}$ are not adjacent to each other, is :

Options:

Answer:

B

Solution:

Total - when $B_{1}$ and $B_{2}$ are together $= 2! (3! 4!) - 2! (3! (3! 2!)) = 144$

TestHub · Accepted Answer

Total - when B 1 ​ and B 2 ​ are together = 2 ! ( 3 ! 4 !) − 2 ! ( 3 ! ( 3 ! 2 !)) = 144

Mathematics - Permutation Combination Question with Solution | TestHub

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