Hi sana.noor,

This question can be solved in a couple of different ways, but I'm going to approach it without any special formulas. You need to be willing to put some work on the pad and figure things out, but the math involved is not crazy.

First, how many different ways can you get 3 numbers to add up to 16 (the prompt says that you can use duplicate numbers)?

871
862
853
844
772
763
754
664
655

Now that we've got the "groups" of numbers that are possible, we have to figure out all of the 3-digit numbers that can be made from these groups.

871 can gives us....
178, 187, 718, 781, 817, 871 = 6 numbers
So if we have 3 distinct digits, then we get 6 numbers

If we have a duplicate digit though....
844 can give us...
448, 484, 844 = 3 numbers
So, if we have a duplicate digit, then we only get 3 numbers

871 = 6 numbers
862 = 6 numbers
853 = 6 numbers
844 = 3 numbers
772 = 3 numbers
763 = 6 numbers
754 = 6 numbers
664 = 3 numbers
655 = 3 numbers

Total = 42 numbers

Final Answer: C

GMAT assassins aren't born, they're made,
Rich

sana.noor wrote:A 3-digit integrer consists of digits 1, 2, 3, 4,5,6,7 and8 and the digits can be used more than once. If the sum of the digits is 16, how many such numbers are possible?
a) 32
b) 40
c) 42
D) 56
E) 64

I have used inserting stick or separator technique to solve this problem but i couldnt find the right answer...please help

The stick/separator technique doesn't work here for two main reasons.
First, this question doesn't allow any digits to be zero, and the technique allows for zeros.
Second, restricts the maximum digit value to 8, and the technique allows for greater values.

So, if you were to use this technique, then you'd have to go back and eliminate all of the instances that break the rule that the digits must be between 1 and 8 inclusive (far too much work!!!)

Here's an example of a question in which you can apply the stick/separator technique: https://www.beatthegmat.com/positive-int ... 87765.html

Cheers,
Brent