(A) Let the positions be $1, 2, 3, 4, 5, 6, 7, 8$. If $A$ is at position $i$,then $B$ must be at position $i+3$ (to have two people between them). The

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(A) Let the positions be $1, 2, 3, 4, 5, 6, 7, 8$. If $A$ is at position $i$,then $B$ must be at position $i+3$ (to have two people between them). The possible pairs of positions $(i, i+3)$ are $(1, 4), (2, 5), (3, 6), (4, 7), (5, 8)$. There are $5$ such pairs. Since $A$ and $B$ can interchange their positions,there are $5 \times 2 = 10$ ways to place $A$ and $B$. The remaining $6$ people can be arranged in the remaining $6$ positions in $6!$ ways. Total ways $= 10 \times 6! = 10 \times 6 \times 5! = 60 \times 5!$.

Class 11 Mathematics Permutation and Combination Definition of permutation, Number of permutations with or without repetition, Conditional permutations

Medium

In how many ways can $8$ people stand in a line such that there are always exactly two people between two specific people $A$ and $B$?

A
$60 \times 5!$
B
$15 \times 4! \times 5!$
C
$4! \times 5!$
D
None of these

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Permutation and Combination

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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 $8$ people stand in a line such that there are always exactly two people between two specific people $A$ and $B$?

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