Question

A tea party is arranged for $16$ people along two sides of a long table with $8$ chairs on each side. $4$ men wish to sit on one particular side and two on the other side. The number of ways that they be seated is :

• A. ${}^{10}{C}_4$
• B. $8!$
• C. ${}^{10}{C}_4\times 8!$
• D. ${}^{10}{C}_4\times (8!{)}^2$

Answer 1

The correct option is D ${}^{10}{C}_4\times (8!{)}^2$

$16$ people

$4$ wish to sit on one side (Lets say Left side)

$2$ wish to sit on other side (Lets say right side)

no. of ways of arranging $4$ people =we have select $4$ out of $8$ seats and arrange these $4$ people these.

${\Rightarrow }^8{C}_4\times 4!$ (Left side)

$i)$ no of ways of arranging $2$ people=

select $2$ out of $8$ seats and arrange $2$ people these.

${\Rightarrow }^8{C}_2\times 2!$ (Right side)

$ii)$ Of the remaining $4$ seats on the side we select $4$ people out of rearrange $10$ people and rearrange these.

${\Rightarrow }^{10}{C}_4\times 4!$

$iii)$ there are $6$ people left and have to be rearranged in $6$ empty seats on right side.

$\Rightarrow 6!$

Total no. of ways ${\Rightarrow }^8{C}_4\times 4!{\times }^8{C}_2\times 2!{\times }^{10}{C}_4\times 4!\times 6!$

$\Rightarrow \frac{8!}{4!4!}\times 4!\times \frac{8!}{6!2!}{\times }^{10}{C}_4\times 4!\times 6!$

${\Rightarrow }^{10}{C}_4(8!{)}^2$