JEE-Mains 2015 (6/30)

Firstly, all permutations of $3, 5, 6, 7, 8$ will yield $5$ -digit integers, and they will be larger than $6000$ . That is already $5!$ integers. However, there are also $4$ digit numbers larger than $6000$ . Those will begin with either a $6$ , $7$ or $8$ . For each of these, there are $4!$ ways to rearrange the numbers afterwards, so our total is $5!+4! \times 3$ .

Solved it without even leaving the flat-Sherlock, Series 3, The Empty Hearse.

$number$ $of$ $4-digit$ $integers$ $that$ $is$ $greater$ $than$ $6000$ $is$

$N_1=(3)(4)(3)(2)=72$

$number$ $of$ $5-digit$ $integers$ $that$ $is$ $greater$ $than$ $6000$ $is$

$N_2=(5)(4)(3)(2)=120$

$N=72+120=192$