Hi,

A five digits number has to be formed by using the digits 1, 2, 3, 4 and 5 without repetition such that the even digits occupy odd places. Find the sum of all such possible numbers.

Means even digits which are 2 and 4 (out of these five given numbers), and can come at position of unit's or hundreads' or ten-thousand's position (which are odd places when we consider the number from right to left...). Out of these three positions, at any two positions 2 and 4 will come in [tex]{3 \choose 2}[/tex]

= 3 ways and 2 and 4 can be permuted in 2! ways. Remaining 3 digits (1, 3 & 5) will come in the remain positions in 3! ways. So in all there will be 3 x 2! x 3! = 36 numbers. And the sum of these 36 numbers is to be evaluated without simply adding all those numbers (which is not a method).....

TIA.