Question

There are $10$ seats in the first row of a theatre of which $4$ are to be occupied. The number of ways of arranging $4$ persons so that no two persons sit side by side is:

• A. ${{}^{7}C}_4$
• B. $4.{{}^{7}P}_3$
• C. ${{}^{7}C}_{3}4!$
• D. $840$

Answer 1

The correct option is D $840$

$\underset{–}{x}1$ $\underset{–}{y}1$ $\underset{–}{z}1$ $\underset{–}{w}1\underset{–}{u}$

$-$ represents empty seat

represents one person

$x+1+y+1+2+1+w+1+u=10$

$x+y+z+w+u=6$

$x,u\ge 0$ and $y,z,w>0$

$\therefore x+y+z+w+u=3$

$x,y,z,w,u\ge 3$

Number of ways $=\frac{(5+(3-1\left)\right)!}{(5-1)\left(3\right)!}$

$=\frac{7!}{4!3!}=\frac{7!}{4!3!}$

$\therefore$ total number of arrangements

$=\frac{7!}{4!3!}.4!\to$ Arrange of people

${=}^7{C}_4.4!$

$=840$