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:

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: 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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