There are 9 chairs in a room on which 6 perso | vedclass

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(A) Total chairs = $9$. Total persons = $6$.One person (the guest) has a specific chair reserved for them. This means the guest has only $1$ way to be

Vedclass · Accepted Answer

$6720$

Vedclass · Answer

(A) Total chairs = $9$. Total persons = $6$.One person (the guest) has a specific chair reserved for them. This means the guest has only $1$ way to be seated.After the guest is seated,there are $5$ persons remaining and $8$ chairs remaining.The number of ways to seat the remaining $5$ persons on the remaining $8$ chairs is given by the permutation formula $P(n, r) = \frac{n!}{(n-r)!}$.Here,$n = 8$ and $r = 5$.Number of ways = $P(8, 5) = \frac{8!}{(8-5)!} = \frac{8!}{3!} = 8 	imes 7 	imes 6 	imes 5 	imes 4 = 6720$.Therefore,the total number of ways is $1 	imes 6720 = 6720$.

There are $9$ chairs in a room on which $6$ persons are to be seated. Out of these $6$ persons,one is a guest who must sit on one specific chair. In how many ways can they be seated?

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