Start with the option which is the least in quantity i.e., 5.

With 5 colors, we can get 5 clients with single colors and for clients with 2 color we can have 5C2 colors ( order is not important here)
= 5*4 / (1*2) = 10 colors.
So, total colors possible = 10+ 5 = 15.

This is sufficient to cover the 12 clients. Hence, answer is E!

You can cross check by checking with 4 colors which will give 10 possible combinations (not sufficient to cover 12 clients)

nafiul9090 wrote:John has 12 clients and he wants to use color coding to identify each client. If either a single color or a pair of two different colors can represent a client code, what is the minimum number of colors needed for the coding? Assume that changing the color order within a pair does not produce different codes.

24
12
7
6
5

OA is E

is there any shortcut approach to attack this type of question ??

Say no. of colours used are n. So by single colour, total codes are n and with 2 colours codes will be nC2.

=> 12 < n + nC2

12 < n + n.(n-1)/1.2

24 < n^2 + n

24 < n.(n+1)

as we see LHS is multiple of 2 consecutive integers hence n = minimum 5.