(A) First,arrange the $5$ girls in a row. This can be done in $5!$ ways.There are $6$ possible gaps created between and at the ends of the girls,repre

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(A) First,arrange the $5$ girls in a row. This can be done in $5!$ ways.There are $6$ possible gaps created between and at the ends of the girls,represented by $\times$:$\times G_1 \times G_2 \times G_3 \times G_4 \times G_5 \times$To ensure no two boys are together,we must place the $3$ boys in these $6$ available gaps.This can be done in $^6P_3$ ways.Total number of ways $= 5! \times ^6P_3 = 120 \times (6 \times 5 \times 4) = 120 \times 120 = 14400$.

Class 11 Mathematics Permutations and Combinations Definition of permutation, Number of permutations with or without repetition, Conditional permutations

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In how many ways can $5$ girls and $3$ boys be seated in a row so that no two boys are together?

A
$14400$
B
$14400$
C
$14400$
D
$14400$

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Permutations and Combinations

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Definition of permutation, Number of permutations with or without repetition, Conditional permutations

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In how many ways can $5$ girls and $3$ boys be seated in a row so that no two boys are together?

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