Let S=(1,2,3,4,5), How would can I list all the 3 permutations of S and all the 3 combinations of S?

Answer 1

Hey, you can't expect to do math and be afraid of a little tedium.

There are 60 permutations of 3 things taken in order from S. Think 5 choices for the first, 4 for the second (can't math the first), 3 for the third (can't match first or second), giving 5*4*3 = 60.

The list is straightforward if you do it in a reasonable order.
123, 124, 125, 132, 134, 135, 142, 143, 145, 152, 153, 154 are the 12 permutations that start with 1. There are 12 that start with each other number. 213, 214, 215, 231, 234, 235, 241, 243, 245, 251, 253, 254, and you can do the rest.

Combinations means that the order doesn't matter. For example, 123, 132, 213, 231, 312, and 321 all correspond to the same combination {123}. In fact, each of these combinations corresponds to 6 permutations, which means there are 60 / 6 = 10 combinations. That's also what happens when you compute C(5,3) = 5! /(3! 2!). Here they are:
{123}, {124}, {125}, {134}, {135}, {145}, {234}, {235}, {245}, {345}

Answer 2

There are 5P3 = 60 permutations taken 3 at a time and

5C3 = 5P3/3! = 10 combinations.

Surely you're not expected to list the 60 subsets of 5P3.