B Combinations of 6 taken 4 at a time

[Vector1962]

TL;DR

the calculation 6C4 shows 15 but what if all sets are to be distinct?

the calculation 6C4 shows 15 but what if all sets are to be distinct? meaning 1,2,3,4 is the same as 4,3,2,1. I made a tree diagram and i get 10... assuming i did that correctly...?

 

[PeroK]

Summary:: the calculation 6C4 shows 15 but what if all sets are to be distinct?

the calculation 6C4 shows 15 but what if all sets are to be distinct? meaning 1,2,3,4 is the same as 4,3,2,1. I made a tree diagram and i get 10... assuming i did that correctly...?

There are definitely 15 distinct combinations. It's easier to count all the ways of leaving two numbers out.

 

[FactChecker]

6C4 is the number of combinations, meaning that order does not matter. 6P4 is the number of permutations, meaning that order does matter.
6C4 = 6!/(4!*2!) = 30/2=15.
6P4 = 6!/2! = 720/2 = 360.
The extra 4! in the denominator of 6C4 divides by the number of ways that the 4 selected can be ordered, so the result is the number of possibilities ignoring their order.