Combination Concept And Shortcut Methods

With combinations, things are pretty much relaxed.

Let’s go back to the example about a competition with eight contestants. When dealing with combinations it doesn’t matter who gets what prize. As a matter of fact, the prizes themselves do not even matter. We just want to give out 3 prizes to three people out of eight.

Combinations will thus tell us the number of different ways that the 3 prizes can be distributed randomly among the 8 contestants.

First, Let’s say we give the 3 prizes to three people Irene, Jason and Bob in no particular order. There will be 3 options for the first person 2 options for the second and only 1 option for the last. This means that we’ll have 3.2.1= 6 ways in which the prizes may be given to the 3 people.

But what happens if we have 8 people and we want to give prizes to only three of them? How many different ways can the 3 prizes be distributed among the 8 people if each prize goes to only one person.

We’ll find the total number of options available (i.e. permutations=336)
and then eliminate redundancies by divided by the number of ways that 3 prizes can be distributed among 3 people [i.e. 6 ways (3!) ].
Hence there are 336/6 = 56 ways in which 3 prizes can be randomly distributed among 8 people.

For combinations, we generally find all possible permutations and then remove redundancies by dividing with a factorial.

Therefore, the combination formula is derived from finding all possible ways of selecting k items from a set of n items and then dividing the result by k factorial (k!)

When put into perspective we see that:

Combination Concept