Question

$7$ gentlemen and $4$ ladies can sit at a round table so that two particular ladies may not sit together in
(a) $7!3!2!{\text{ ways}}$
(b) $6! {}^6{P_4}$ ways
(c) $6! {}^7{P_4}$ ways
(d) $6! {}^4{P_2}$ ways

Answer 1

Hint: This problem can be solved by permutations. A Permutation of a set is an arrangement of its members into a sequence or linear ordered, a rearrangement of its elements.

Here we have to arrange $7$ gentlemen and $4$ ladies can sit at a round table so that two particular ladies may not sit together.
We know that $n$ things can be arranged around a round table in $\left( {n - 1} \right)!$ ways
First, we arrange $7$ gentlemen at a round table in $6!$ ways
Then we arrange $4$ ladies so that two particular ladies may not sit together.
We know that if $m$ gentlemen and $n$ ladies are to be seated at a round table so that two particular ladies may not sit together can be arranged in ${}^m{P_n}$ ways.
There are $7$ gaps between men and $4$ ladies are to be placed.
By using the above formula this can be done in ${}^7{P_4}$ ways.
Therefore, that total arrangement can be done in $6!{}^7{P_4}$ ways.
Thus the answer is option (c) $6!{}^7{P_4}$ ways.

Note: In this problem we have used multiplicative principle of permutation i.e. if there are $x$ ways of doing one thing and $y$ ways of doing another, then the total number of ways of doing both the things is $xy$ ways.