Question

There are 9 chairs in a room on which 6 persons are to be seated, out of which one is the guest with one specific chair. In how many ways can they sit?
(a) 6720
(b)60480
(c) 30
(d) 346

Answer 1

We have to use combinations here. According to the given condition, one specific chair is for the guest. Therefore, the remaining chairs will be 8 and the remaining persons will be 5. Now, we have to choose 5 chairs from 8. We can do this in $^{8}{{C}_{5}}$ . The 5 persons can be arranged in 5! ways. Therefore, the required number of ways can be found by multiplying $^{8}{{C}_{5}}$ with 5!.

Complete step by step answer:
We are given that there are 9 chairs in a room and 6 persons have to be seated. Of these, one is the guest with one specific chair. Therefore, the remaining chairs will be 8 and the remaining persons will be 5.
We know that the r number of elements can be selected from n in $^{n}{{C}_{r}}$ ways. Therefore, we can choose 5 chairs from 8 in $^{8}{{C}_{5}}$ ways.
Now, we can arrange 5 persons in 5! ways.
Therefore, the number of ways of arranging 5 persons on 8 seats can be found as follows.
${{\Rightarrow }^{8}}{{C}_{5}}\times 5!$
We know that $^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ . Therefore, the above expression can be written as
$\begin{aligned} & \Rightarrow \dfrac{8!}{\left( 8-5 \right)!5!}\times 5! \\\ & =\dfrac{8!}{3!5!}\times 5! \\\ \end{aligned}$
Let us cancel the common terms.
$\begin{aligned} & \Rightarrow \dfrac{8!}{3!\require{cancel}\cancel{5!}}\times \require{cancel}\cancel{5!} \\\ & =\dfrac{8!}{3!} \\\ \end{aligned}$
We know that $n!=n\times \left( n-1 \right)!$ . Therefore, we can expand the above expression as
$\Rightarrow \dfrac{8\times 7\times 6\times 5\times 4\times 3!}{3!}$
Let us cancel the common terms.
$\begin{aligned} & \Rightarrow \dfrac{8\times 7\times 6\times 5\times 4\times \require{cancel}\cancel{3!}}{\require{cancel}\cancel{3!}} \\\ & =8\times 7\times 6\times 5\times 4 \\\ & =6720 \\\ \end{aligned}$

So, the correct answer is “Option a”.

Note: Students must be thorough with the formulas of factorials, permutations and combinations. They have a chance of making a mistake by writing the formula of the combination as $\dfrac{n!}{\left( n-r \right)!}$ which is the formula of permutation. Students have a chance of making a mistake by adding $^{8}{{C}_{5}}$ and 5! To find the required number of ways.