First slide

Permutations

Question

Three boys of class X, four boys of class XI, and five boys of class XII sit in a row. The total number of ways in which these boys can sit so that all the boys of same class sit together is equal to

Moderate

A

$(3!)^{2} (4!) (5!)$
B

$(3!) (4!)^{2} (5!)$
C

$(3!) (4!) (5!)$
D

$(3!) (4!) {(5!)}^{, 2}$

Solution

We can think of three packets. one consisting of three boys of class X, other consisting of 4 boys of class XI and last one consisting of 5 boys of class XII. These packets can be arranged in it ways and contents of these packets can be number arranged in 31 4! and 5 ! ways, respectively. Hence, the total number of ways is 3 !x 3 ! x 4! x 5 !

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