Question

Three letters are chosen at random from the letters P, A, N, S, O, L. What is the size of the sample space of this experiment? Assume that the order of the letters matters. That is, PAN is different from APN

Answer 1

You need to find the number of permutations of the 6 different letters, taken 3 at a time.
$6P3=\frac{6!}{(6-3)!}=\frac{6\times5\times4\times3\times2\times1}{3\times2\times1}=120$
Therefore the sample space is 120 possible outcomes.

Answer 2

If all the letters are different,
Using combinations:

6C3 =6! /3!* (6-3)!
Simplifying,
6C3 = 6x5x4 /1x2x3
       = 120/6
        = 20 ways 
Hence,
The size of the sample space will be 20.