Question

The no. of $10$ letter codes that can be formed using the characters $x,y,z,r$ with the restriction that $x$ appears exactly thrice and $y$ appears exactly twice in each such codes is

• A. $60840$
• B. $88400$
• C. $80640$
• D. $64080$

Answer 1

The correct option is C

$80640$

Explanation for the correct option.

Finding the no. of $10$ letter codes that can be formed using the characters $x,y,z,r$ with the restriction that $x$ appears exactly thrice and $y$ appears exactly twice in each such codes is

Given: $x,y,z,r$

$x$ appears exactly thrice and $y$ appears exactly twice

$x$	$y$	$z$	$r$	Number of codes
$3$	$2$	$0$	$5$	$\frac{10!}{(3!.2!.5!)}=2520$
$3$	$2$	$1$	$4$	$\frac{10!}{(3!.2!.1!.4!)}=12600$
$3$	$2$	$2$	$3$	$\frac{10!}{(3!.2!.2!.3!)}=25200$
$3$	$2$	$3$	$2$	$\frac{10!}{(3!.2!.3!.2!)}=25200$
$3$	$2$	$4$	$1$	$\frac{10!}{(3!.2!.4!.1!)}=12600$
$3$	$2$	$5$	$0$	$\frac{10!}{(3!.2!.5!.0!)}=2520$

Therefore, total number of codes $=2520+12600+25200+25200+12600+2520=80640$

Hence the correct answer is option (C).