Question

Two classrooms $A$ and $B$ having capacity of $25$ and $(n-25)$ seats respectively. $A_n$ denotes the number of possible seating arrangements of room ${}^{\prime }{A}^{\prime }$, when ${}^{\prime }{n}^{\prime }$ students are to be seated in these rooms, starting from room ${}^{\prime }{A}^{\prime }$ which to be filled up full to its capacity. If $A_n-A_{n-1}=25!{(}^{49}{C}_{25})$ then ${}^{\prime }{n}^{\prime }$ equals.

• A. $50$
• B. $48$
• C. $49$
• D. $51$

Answer 1

Answer: (A)

$50$

$\therefore A_n{=}^n{C}_{25}\times 25!$

$A_{n-1}{=}^{n-1}{C}_{25}\times 25!$

$\therefore A_n-A_{n-1}{=}^n{C}_{25}\times 25!{-}^{n-1}{C}_{25}\times 25!$

$=25!{(}^n{C}_{25}{-}^{n-1}{C}_{25})=25!{(}^{49}{C}_{25})$

${\Rightarrow }^n{C}_{25}{-}^{n-1}{C}_{25}{=}^{49}{C}_{25}$

$\Rightarrow \frac{n!}{25!(n-25)!}-\frac{(n-1)!}{25!(n-26)!}=\frac{49!}{(25!)(24!)}$

$\Rightarrow \frac{(n-1)!}{(n-26)!}[\frac{n}{n-25}-1]=\frac{49!}{24!}$

$\Rightarrow \frac{(n-1)!}{(n-26)!}\left[\frac{25}{n-25}\right]=\frac{49!}{24!}$

$\therefore$ By comparing we get

$n=50$