1) 2137^754=(2130+7)^754

on binomial expansion,we find that the first teram is not divisible by 10 while others are divisible by 10..

SO only first term will contribute to last digit..

the first term is 7^754

the last digit of this will be the last digit of given question.'

now observe the last digit in following arguments.

7^1=7, 

7^2=49,

7^3=513

7^4=last digit being 1

and then the last digits repeat

the last digits are 7,9,3,1,7,9,3,1,

SO 7^(4n+1)---last digit is 7

7^(4n+2)---- last digit is 9

..

so on dividing 754 by 4 we get 2 as remainder.........

so the last digit will be 9.......

2) there are 5P4 ways of forming a 4 digit number with each digit appearing at most once.Consider first the one's place. In those 120 numbers, 1,2,3,4, and 5 appear exactly the same number of times, that is, each digit appears 120/5 = 24 times.

For the tens place, same thing: 1,2,3,4,and 5 appear equally the same number of times. Each digit appears 24 times.

The argument for the hundreds and thousands places are the same.

Thus the sum of all those distinct 120 numbers is
1,000*24*(1+2+3+4+5)+ 100*24*(1+2+3+4+5) + 10*24*(1+2+3+4+5) + 1*24*(1+2+3+4+5)

= 1,111*24*(1+2+3+4+5)

= 1,111*24*15

= 3,99,960 i.e option (B) 

PLEASE APPROVE !

3) HINT : Use binomial expansion of (bc + a (b + c))6.

answer is 3 . 6!/2!3! = 720/12 = 60

Please approve !

hey sudeesh what about the other question , you have attempted the easiest one; i have posted this questions thinkinkg that some one will answer the tough questions but ......

dear sudheesh,

see this method you should solve like this,well your attempt was good.i'm enclosing you with solution of 19th question also.