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The alphabets in the sorted order will be $\{C, E, I, K, R, T\}$ .

We could replace the sorted set with sorted numbered set $\{1, 2, 3, 4, 5, 6\}$ .

Hence the process of generating $7$ -letter words can be transformed into the problem of generating $7$ -digit numbers from the above set.

The position of $1531426$ (CRICKET) in sorted order would be as follows-

For the number $1531426$ :

$7^{th}$ digit is the smallest possible.

$6^{th}$ digit came after $\{1, 2, 3, 4\}$ , hence after $4 \times 5!$ numbers.

$5^{th}$ digit came after $\{1, 2\}$ , hence after $2 \times 4!$ numbers.

$4^{th}$ digit is the smallest possible.

$3^{rd}$ digit came after $\{2\}$ , hence after $2!$ numbers.

$2^{nd}$ digit is the smallest possible.

$1^{st}$ digit is the smallest possible.

Hence position of $1531426$ (CRICKET) $=$

Position of numbers before $1531426$ (CRICKET) $+ 1 =$

$4 \times 5! + 2 \times 4! + 2! + 1 = 531$

this question is from tata mcgraw hill this

A normal JEE type question. also important for exams like CAT. With CC->5! ,similarly 5! With CE,CI,CK each.total till here =480.Now 4! With CRC & CRE each and 2! With CRICE.Next is CRICKET!