Combination: Choosing Without Ordering
Imagine you have three friends: A, B, and C. You need to pick two of them to form a committee. How many different committees can you make?
You could pick A and B, or A and C, or B and C. That's three committees. Notice something crucial: the committee {A, B} is exactly the same as {B, A}. Who gets picked first doesn't matter — only who is on the committee matters. This is the core idea of a combination: a selection where the order of selection does not matter.
The Intuition: "How many groups?"
A combination answers the question: "From a set of n distinct items, how many ways can I choose r of them, ignoring the order in which I pick them?"
Think of it as forming a team, a committee, a hand of cards, or a set of ingredients for a recipe. In all these cases, swapping two members around doesn't give you a new team — it's still the same group.
The Precise Statement
A combination is a selection of r objects from a set of n distinct objects, where the order of selection does not matter. The number of such combinations is denoted by:
(rn)ornCr
and is given by the formula:
(rn)=r!(n−r)!n!
where n! (read "n factorial") means n×(n−1)×(n−2)×⋯×2×1, and 0!=1 by definition.
Why the r! in the denominator?
This is the key to understanding combinations. If order did matter, the number of ways to pick r objects from n would be n×(n−1)×⋯×(n−r+1), which is written as (n−r)!n!. That's called a permutation.
But in a combination, every group of r objects can be rearranged in r! different orders. Since we don't care about order, we divide by r! to "collapse" all those orderings into one single group.
So:
(rn)=number of ways to order each selectionnumber of ordered selections=(n−r)!n!÷r!=r!(n−r)!n!
A Quick Example
From a class of 10 students, how many ways can you choose a team of 3?
Here n=10, r=3.
(310)=3!7!10!=3×2×110×9×8=6720=120
So there are 120 different teams possible. …