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Question

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(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

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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.

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.