Lecture 24: Combinations
Take a set with elements: .
A combination from is a subset formed by choosing any elements from :
The number of combinations of elements is written .
Example-1
There are players available to our coach of which she can choose to form a team consisting of people.
How many possible choices of teams does she have?
Answer We have a set of size people. What is needed here is a -combination from this set.
The answer is therefore .
Example-2
There are teams in a round-robin tournament where each pair of teams play exactly one match. How many matches need to be played?
Answer Each match is a -combination from the set of teams. Therefore, the number of matches is .
Example-3
A cricket coach can make up a team of cricketers amongst possible candidates. She has
How many possible choices of teams does she have, given her plans above.
Answer .
Example-4
There are 10 stalls in the farmer's market. Each stall sells apples and oranges.
We can (a) skip a stall, (b) buy two apples in each stall or (c) alternatively buy two oranges.
How many ways are there of buying apples and oranges.
Computing the Number of Combinations
So far, we have seen . What exactly is it in terms of numbers? There are two ways of thinking about it:
From Permutations to Combinations
In a -permutation of elements:
The number of -permutations is .
In a -combination, the order of choice does not matter.
Let us relate -permutations and -combinations.
Here is a way of looking at -permutations:
Thinking about things above, we get the following identity: .
We already reasoned out . Therefore, the formula for should be

Let us play around with it:

Claim .
Let us give an intuitive reason for this: means choosing elements out of a total of . We can look at it also as choosing which elements to include or choosing which elements to exclude.
There are ways of choosing elements to include in our subset. On the other hand, there are ways of choosing which elements to exclude. Both views are counting the same
thing. Therefore they better be the same.
Mathematically, just write down the formulae:

Example-1
Six people would like to sit around a round table. It does not matter where a person sits, but what matters is who sits next to him/her. In other words, choices
1,2,3,4,5,6 or 6,1,2,3,4,5 are the same since each choice leaves the same neighbors. How many possible seating choices are there?
Answer Let us say the number is . We see that once we choose a seating order, we can obtain different permutations of it that are equivalent since they correspond to the
same seating order. Therefore .
Recursively Counting Combinations
Here is another way of thinking. We need to choose a subset of elements from a set of size : .
-
If then we have no choice. If then we have choice (the empty set) and if then we have choices.
Otherwise, choose whether or not to include the smallest numbered element .
Therefore

Connection between and the Pascal's triangle in class.
Binomial Theorem
Now for a bit of algebra. Let us consider polynomials .
.
.
.
.
(answer in class).
The question is what should be.
Take . I am interested in the coefficients of various terms.
Let us start with .
How many ways are there of forming ? Simple, we have to choose 2 s (the remaining will naturally yield ). There are four possible multiplicands of which
we can choose “x” in ways.
If you understood the argument above then you have understood the theorem we are about to prove.
Binomial Theorem
For all natural numbers , we can expand
as

We can write it succinctly as .
Proof What is the coefficient of in .
We write as .
There are multiplicands.
To form a term where , we can choose out of multiplicands to “supply” the . The remaining will be chosen automatically to supply the s.
There are choices. Each choice appears as a term which are then collected up.
Our reasoning above shows that the coefficient of where should be .
Example-1
Expand ?
Answer .
Applications of Binomial Theorem
We can use the Binomial theorem to show some properties of the function.
1. .
Proof: Take the expansion of and substitute .

2. Let be an even number. Then we have .
Proof: Take and set . Write a similar result for odd.
3. Evaluate: .
Proof: Take . Using binomial theorem, we have
.
Take the derivative of both sides w.r.t . We get:

Substituting , we get the required summation to be .
4. Likewise, can you evaluate this summation:

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