### 3.6 - The Binomial and Multinomial Theorems

We have previously learned that a binomial is an expression that contains 2 terms and a multinomial is any expression that contains more than 1 term (so a binomial is actually a special case of a multinomial). We have also previously seen how a binomial squared can be expanded using the distributive law. For example
If we want to cube a binomial or raise it to an even higher power then we could do this by repeatedly applying the distributive law to lower powers. For example
But for very high powers this procedure is very laborious. The binomial theorem provides a shortcut. Similarly, the multinomial theorem provides a shortcut to expanding a multinomial raised to any positive integer power.

### The Binomial Theorem

Take a look at these expansions of a binomial raised to power n, with n = 0, 1, 2, 3, 4 and 5.
Let’s look for patterns in these expansions. Notice the following:
• In each expansion there are n + 1 terms. (For example the bottom (n = 5) expansion has 6 terms.)

• The powers of x start at n and decrease by 1 in each term until they reach 0.

• The powers of y start at 0 and increase by 1 until they reach n.

• The coefficients in each expansion add up to 2n. (For example in the bottom (n = 5) expansion the coefficients 1, 5, 10, 10, 5 and 1 add up to 25 = 32.)

• The coefficients are exactly the numbers that appear in Pascal’s Triangle, shown here:
What is Pascal’s Triangle? It is an infinitely tall triangle of numbers that can be constructed as follows:

• The first number and the last number in each row is a 1.

• Every number in the interior of the triangle is the sum of the two numbers above left and above right, like this:
• Pascal’s Triangle above is animated to show how each row is generated from the row above.

• Each number in Pascal’s triangle can also be generated using the so-called combination function as shown here:
Pascal’s triangle is really just a listing of all the possible values of the combination function. The combination function is defined as
where n! or “n factorial” is
The combination function gives the number of distinct groups of k objects that can be chosen from n distinct objects when the order in which they are chosen doesn’t matter.

Here are some examples of calculations that use the combination function:

Example: From 4 people, how many distinct groups of 2 people be chosen to work on some project?

Solution: The answer is
If the persons are called A, B, C and D then the distinct groups are AB, AC, AD, BC, BD and CD.

Example: In the expansion of (a+b)4 what is the coefficient of the term a3b?

Solution: The answer is 4. The reason is that the term a3b comes from multiplying a b from inside 1 of the brackets and a’s from inside the remaining 3 brackets. There are
ways of choosing which bracket the b should come from. The blue arrows show one of these ways:

Putting all of this together gives the binomial theorem, which states that a power of a binomial can be expanded like this:
Using summation notation the binomial theorem can also be written like this:

### The Multinomial Theorem

The multinomial theorem extends the binomial theorem. It describes the result of expanding a power of a multinomial. We will show how it works for a trinomial. Multinomials with 4 or more terms are handled similarly.

Consider (a + b + c)4. The brute force way of expanding this is to write it as

(a + b + c) (a + b + c) (a + b + c) (a + b + c),
then apply the distributive law, and then simplify by collecting like terms. After distributing, but before collecting like terms, there are 81 terms. (This is because every term in the first brackets has to be multiplied by every term in the second brackets, giving 9 terms. Each of these has to be multiplied by every term in the third brackets, giving 27 terms. Finally each of these has to be multiplied by every term in the fourth brackets, giving 81 terms.) Many of the terms look different before simplifying, but are identical after simplifying. For example the four terms abbb, babb, bbab and bbba (where the a comes from either the first, second, third or fourth brackets) can be simplified and collected to give 4ab3. Thus to understand the various coefficients in the final result shown below we need to consider what distinct terms can occur and how many ways there are of getting each of them:
Notice the following:
• There are 15 distinct terms.

• Each term is of degree 4.

• The coefficients add up to 34 = 81 (As mentioned above this is the number of terms before collecting like terms.)
We will construct this expansion in steps, and in so doing, derive the multinomial theorem.

1. Start with three nested summations:
The problem is that this triple summation produces 5·5·5=125 terms of various degrees ranging from a0b0c0 to a4b4c4.

2. Introduce a filter factor that equals 1 for the wanted terms (those of degree 4) and 0 for the unwanted terms. This is the Kronecker delta function, denoted δi, j and defined as:
Now change the summation to read
It contains only the 15 wanted terms of degree 4. The only remaining problem is that every coefficient equals 1.

3. Construct the correct coefficient. This is the so-called multinomial coefficient:
This multinomial coefficient gives the number of ways of depositing 4 distinct objects into 3 distinct groups, with i objects in the first group, j objects in the second group and k objects in the third group, when the order in which they are deposited doesn’t matter.

For example the coefficient of the a1b1c2 term uses i = 1, j = 1 and k = 2 and equals
With this coefficient the expansion reads

We can generalize this to give us the nth power of a trinomial. (Just change all the 4’s to n’s.)
This is the multinomial theorem for 3 terms. It easily generalizes to any number of terms. For example the multinomial theorem for 4 terms reads:

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