(A) From a committee of $8$ persons,a chairman and a vice chairman are to be chosen such that one person cannot hold more than one position.Here,the n

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(A) From a committee of $8$ persons,a chairman and a vice chairman are to be chosen such that one person cannot hold more than one position.Here,the number of ways of choosing a chairman and a vice chairman is the permutation of $8$ different objects taken $2$ at a time.Thus,the required number of ways $= ^{8}P_{2} = \frac{8!}{(8-2)!} = \frac{8!}{6!} = \frac{8 \times 7 \times 6!}{6!} = 8 \times 7 = 56$.

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

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From a committee of $8$ persons,in how many ways can we choose a chairman and a vice chairman assuming one person cannot hold more than one position?

A
$56$
B
$42$
C
$28$
D
$16$

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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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From a committee of $8$ persons,in how many ways can we choose a chairman and a vice chairman assuming one person cannot hold more than one position?

Similar Questions

The number of ways in which $5$ girls and $3$ boys can be seated in a row so that no two boys are together is:

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