Question

Let $P_n$ denotes the number of ways in which three people can be selected out of ${}^{\prime }{n}^{\prime }$ people sitting in a row, if no two of them are consecutive. If $P_{n+1}-P_n=15$ then the value of ${}^{\prime }{n}^{\prime }$ is :

• A. $7$
• B. $8$
• C. $9$
• D. $10$

Answer 1

The correct option is C $9$

We have $n$ people and we have to choose $3$. Let $A,B,C$ be those person

So,

$x_1$ denotes number of to left of $A$

$x_2$ denotes number of to left of $A$ and $B$

$x_3$ denotes number of to left of $B$ and $C$

$x_4$ denotes number of to left of $B$

$x_1+x_2+x_3+x_4=n-3$

As $A,B,C$ cannot be consecutive , hence

$x_2,x_3\ge 1$

$x_1,x_4\ge 0$

Replace $x_2=x_2^{\prime }+1$ and $x_3=x_3^{\prime }+1$ so that

$x_2^{\prime },x_3^{\prime }\ge 1$

$x_1+x_2^{\prime }+1+x_3^{\prime }+1+x_4=n-3$

$x_1+x_2^{\prime }+x_3^{\prime }+x_4=n-5$

AS $x_1,x_2^{\prime },x_3^{\prime },x_4\ge 0$ so now we can apply formula for number of non-negative integral solution.

$P_n{=}^{n-5+4-1}{C}_{4-1}{=}^{n-2}{C}_3$

$P_n{=}^{n-2}{C}_3$

$P_{n-1}{=}^{n-3}{C}_3$

As per question

$P_n-P_{n-1}=15$

${}^{n-2}{C}_3{-}^{n-3}{C}_3=15$

$\frac{(n-2)(n-3)(n-4)}{6}-\frac{(n-3)(n-4)(n-5)}{6}=15$

$(n-3)(n-4)[n-2-n+5]=90$

$(n-3)(n-4)=30$

$(n-3)(n-4)=6\times 5$

$n-3=6$

$n=9$