Multinomial theorem: The binomial theorem primarily helps to find the expansion of the form \((x+y)^{n}\). Finding the value of \((x+y)^{2},(x+y)^{3},(x+y)^{4}\) and \((a+b+c)^{2}\) is easy as the expressions can be multiplied by themselves based on the exponent. On the other hand, finding the expansion of \((x+y)^{20}\) or expressions with higher powers. This is where the binomial theorem comes into play.

Similarly, if the expression has more than two terms with higher powers, do you think we can still use the binomial theorem? In such cases, we use the multinomial theorem to expand. In this article, let us learn more about the multinomial theorem and its general term.

What is a Multinomial?

An algebraic expression having two or more (unlike) terms is called a multinomial.

Example:

• \(5 x^{2}+3 x\) is a multinomial with two terms
• \(5 x^{3}-2 x y+7 y^{2}\) is a multinomial with three terms
• \(7 x y-9 y z+6 z x-7\) is a multinomial with four terms

Let us describe a few examples of how to expand a multinomial of exponent \(2\).

Example: Expand \((x+y+z)^{2}\)

We can expand, \((x+y+z)^{2}\) as \((x+y+z)(x+y+z)\)

\(=x(x+y+z)+y(x+y+z)+z(x+y+z)\)

\(=x^{2}+x y+x z+y(x+y+z)+z(x+y+z)\)

\(=x^{2}+x y+x z+x y+y^{2}+y z+z(x+y+z)\)

\(=x^{2}+x y+x z+x y+y^{2}+y z+x z+y z+z^{2}\)

\(=x^{2}+2 x y+2 x z+y^{2}+2 y z+z^{2}\)

Since it is a tedious process to find the expansion of \((x+y+z)^{5},(x+y+z)^{6}\) etc., so we use the multinomial theorem to expand such kind of expression having higher powers.

Definition of Multinomial Theorem

The multinomial theorem is used to expand the sum of two or more terms raised to an integer power.

The multinomial theorem provides a formula for expanding an expression such as \(\left(x_{1}+x_{2}+\cdots+x_{k}\right)^{n}\), for an integer value of \(n\). In particular, the expansion is given by

\({\left( {{x_1} + {x_2} + … + {x_k}} \right)^n} = \sum\limits_{{n_1},\,{n_2},\,…,\,{n_k} \geqslant 0} {\frac{{n!}}{{{n_1}!{n_2}!…{n_k}!}}x_1^{{n_1}}x_2^{{n_2}}…x_k^{{n_k}}} \)

Where,

\(n_{1}+n_{2}+\cdots+n_{k}=n\)

\(n\)! is the factorial notation for \(1 \times 2 \times 3 \times \cdots \times n\).

What is the Multinomial Coefficient?

A multinomial coefficient is used to provide the sum of the multinomial coefficient, which is later multiplied by the variables. It represents the multinomial expansion, and each term in this series contains an associated multinomial coefficient.

We know that multinomial expansion is given by,

\({\left( {{x_1} + {x_2} + … + {x_k}} \right)^n} = \sum\limits_{{n_1},\,{n_2},\,…,\,{n_k} \geqslant 0} {\frac{{n!}}{{{n_1}!{n_2}!…{n_k}!}}x_1^{{n_1}}x_2^{{n_2}}…x_k^{{n_k}}} \)

where \(n_{1}+n_{2}+\cdots+n_{k}=n\)

In the above series, \(x\) is used to describe the terms. The \(k\) is used to describe the number of elements in the series. The \(n\) is used to describe the positive integer power to which the series is raised. Therefore, for non-negative integers \(n_{1}, n_{2}, \ldots n_{k}\) such that \(\sum_{i=1}^{k} n_{i}=n\), the multinomial coefficient is given as

\(\left( {\begin{array}{*{20}{c}}n\\{{n_1},{n_2}, \ldots ,{n_k}}\end{array}} \right)=\frac{n !}{n_{1} ! n_{2} ! \ldots n_{k} !}\)

Proof of Multinomial Theorem

Statement: For a positive integer \(k\) and non-negative integer \(n,{\left( {{x_1}+{x_2}+{x_3}+\cdots+{x_k}}\right)^n}={\sum_{{b_1}+{b_2}+\cdots+{b_k}=n}}\left( {\begin{array}{*{20}{c}}n\\{{b_1},{b_2}, \ldots ,{b_k}}\end{array}} \right)\prod _{j = 1}^kx_j^{{b_j}}\), where \(\left( {\begin{array}{*{20}{c}}n\\{{b_1},{b_2}, \ldots ,{b_k}}\end{array}} \right)\) given as \(\left( {\begin{array}{*{20}{c}}n\\{{b_1},{b_2}, \ldots ,{b_k}}\end{array}} \right)=\frac{n !}{b_{1} ! b_{2} ! \ldots b_{k} !}\)

Proof:

We prove this by the method of mathematical induction in \(k\).

Let us check if the multinomial theorem is true for \(k=1\).

Taking \(k=1\), then we get the \(L.H.S. =\left(x_{1}+x_{2}+\cdots+x_{k}\right)^{n}\)

\(\Rightarrow L.H.S. =\left(x_{1}\right)^{n}\)

\(\Rightarrow L.H.S. =x_{1}^{n}\)

Similarly, when \(k=1\), then we get,

\(\sum\limits_{{b_1} + {b_2} + … + {b_k} = n} {\left( {{b_1},\,{b_2},…,{b_k}} \right)\prod _{j = 1}^kx_j^{{b_j}}} \)

\( = \sum\limits_{{b_1} = n} {\left( \begin{gathered} n \hfill \\ {b_1} \hfill \\ \end{gathered} \right)\mathop \prod \limits_{j = 1}^1 x_j^{{b_j}}} \)

From this, we can see, \(b_{1}=n\)

Hence, we get

\(R.H.S. = \sum\limits_{{b_1} = n} {\left( \begin{gathered} n \hfill \\ {b_1} \hfill \\ \end{gathered} \right)\mathop \prod \limits_{j = 1}^1 x_j^{{b_j}}} \)

\( \Rightarrow R.H.S. = \sum\limits_n {\left( \begin{gathered} n \hfill \\ n \hfill \\ \end{gathered} \right){{\left( {{x_1}} \right)}^{{b_1}}}} \)

\(\Rightarrow R.H.S. =\frac{n !}{n !}\left(x_{1}\right)^{n}\)

\(\therefore R.H.S. =x_{1}^{n}\)

Therefore, \(L.H.S=R.H.S\)

Thus, the multinomial theorem is true for \(k=1\).

Let’s assume that the multinomial theorem is true for \(k=m\), where \(m\) is a positive integer.

Thus, we can say that,

\({\left( {{x_1} + {x_2} + {x_3} +  \cdots  + {x_m}} \right)^n} = \sum\limits_{{b_1} + {b_2} +  \cdots  +{b_m} = n} {\left( {\begin{array}{*{20}{c}}n\\{{b_1},{b_2}, \ldots ,{b_m}}\end{array}} \right)} \prod\limits_{j = 1}^m {x_j^{{b_j}}} \)

Now, that we assumed the multinomial theorem is true for \(k=m\), we will see that if it is true for \(k=m+1\).
Substituting \(k=m+1\) in multinomial theorem, then we get the \(L.H.S.\) as

\(\left(x_{1}+x_{2}+x_{3}+\cdots+x_{m}+x_{m+1}\right)^{n}\)

Now, let us assume that, \(x_{m}+x_{m+1}\) is a single term. So, \(b_{m}+b_{m+1}\) will also be a single term.

Let us assume this term to be \(M\). Hence, the number of terms is

\(m+1-1=m\)

Thus, we can write the multinomial theorem as

\({\left( {{x_1} + {x_2} + {x_3} +  \cdots  + {x_{m – 1}} + \left( {{x_m} + {x_{m + 1}}} \right)} \right)^n} = \sum\limits_{{b_1} + {b_2} +  \cdots  + {b_{m – 1}} + M = n} {\left( {\begin{array}{*{20}{c}}n\\{{b_1},{b_2}, \ldots ,{b_{m – 1}},M}\end{array}} \right)} \)

\(=\mathop \prod \limits_{j = 1}^{m – 1} x_{j}^{b_{j}} \cdot\left(x_{m}+x_{m+1}\right)^{M}\)

Now, we can write the expansion of \(\left(x_{m}+x_{m+1}\right)^{M}\) by using the binomial theorem.

\({\left( {{x_m}+{x_{m+1}}}\right)^M}=\sum\limits_{m+M-{b_m}=M}{\left( {\begin{array}{*{20}{c}}M\\{{b_m},M – {b_m}}\end{array}} \right)} x_m^{{b_m}}{x_{m + 1}}M – {b_m}\)

As we already have \(b_{m}+b_{m+1}=M\), so we can say that

\(b_{m+1}=M-b_{m}\)

Hence, we can write \(\left(x_{m}+x_{m+1}\right)^{M}\) as

\(\sum\limits_{{b_m} + {b_{m + 1}} = M} {\left( {\begin{array}{*{20}{c}}M\\{{b_m},{b_{m + 1}}}\end{array}} \right)} x_m^{{b_m}}{x_{m + 1}}{b_{m + 1}}\)

Now, substitute this value in the expansion, then we get

\(\left(x_{1}+x_{2}+x_{3}+\cdots+x_{m-1}+\left(x_{m}+x_{m+1}\right)\right)^{n}\)

\( = \sum\limits_{{b_1} + {b_2} +  \cdots  + {b_{m- 1}}+M=n}{\left( {\begin{array}{*{20}{c}}n\\{{b_1},{b_2}, \ldots ,{b_{m – 1}},M}\end{array}} \right)} \prod\limits_{j = 1}^{m – 1} {x_j^{{b_j}}}  \cdot {\left( {{x_m} +{x_{m+1}}}\right)^M}\)

\(\Rightarrow\left(x_{1}+x_{2}+x_{3}+\cdots+x_{m-1}+\left(x_{m}+x_{m+1}\right)\right)^{n}\)

\( = \sum\limits_{{b_1} + {b_2} +  \cdots  + {b_{m- 1}}+M=n}{\left( {\begin{array}{*{20}{c}}n\\{{b_1},{b_2}, \ldots ,{b_{m – 1}},M}\end{array}} \right)} \prod\limits_{j = 1}^{m-1} {x_j^{{b_j}}} \)

\( \cdot \sum\limits_{{b_m} + {b_{m + 1}} = M} {\left( {\begin{array}{*{20}{c}}M\\{{b_m},{b_{m + 1}}}\end{array}} \right)} x_m^{{b_m}}x_{m + 1}^{{b_{m + 1}}}\)

\(\Rightarrow\left(x_{1}+x_{2}+x_{3}+\cdots+x_{m-1}+\left(x_{m}+x_{m+1}\right)\right)^{n}\)

\( = \sum\limits_{{b_1} + {b_2} +  \cdots  + {b_{m- 1}}+M=n}{\left( {\begin{array}{*{20}{c}}n\\{{b_1},{b_2}, \ldots ,{b_{m – 1}},M}\end{array}} \right)} \mathop \sum \limits_{{b_m} + {b_{m + 1}}}\)

\( = M\left( {\begin{array}{*{20}{c}}M\\{{b_m},{b_{m + 1}}}\end{array}} \right) \cdot \prod\limits_{j = 1}^{m – 1} {x_j^{{b_j}}}  \cdot x_m^{{b_m}}x_{m + 1}^{{b_{m + 1}}}\)

Now, we will find the value of \({\sum _{{b_1} + {b_2} +  \cdots  + {b_{m – 1}} + M = n}}\left( {\begin{array}{*{20}{c}}n\\{{b_1},{b_2}, \ldots ,{b_{m – 1}},M}\end{array}} \right)\)

\( \cdot \sum_ {{b_m}}  + {b_{m + 1}} = M\left( {\begin{array}{*{20}{c}}M\\{{b_m},{b_{m + 1}}}\end{array}} \right)\) by expanding both these sigmas

\(\sum\limits_{{b_1} + {b_2} +  \cdots  + {b_{m – 1}} + M = n}{\left( {\begin{array}{*{20}{c}}n\\{{b_1},{b_2}, \ldots ,{b_{m – 1}},M}\end{array}} \right)} \sum\limits_{{b_m} + {b_{m + 1}} = M}{\left( {\begin{array}{*{20}{c}}M\\{{b_m},{b_{m + 1}}}\end{array}} \right)} \)

\( = \sum\limits_{{b_1} + {b_2} +  \cdots  + {b_{m – 1}} + M = n} {\frac{{n!}}{{{b_1}!{b_2}! \ldots {b_{m – 1}}!M!}}} \sum\limits_{{b_m} + {b_{m + 1}} = M} {\frac{{M!}}{{{b_m}!{b_{m + 1}}!}}} \)

\( = \sum\limits_{{b_1} + {b_2} +  \cdots  + {b_m} + {b_{m + 1}} = n} {\frac{{n!}}{{{b_1}!{b_2}! \ldots {b_{m – 1}}!M!}}}  \cdot \frac{{M!}}{{{b_m}!{b_{m + 1}}}}\)

\( = \sum\limits_{{b_1} + {b_2} +  \cdots  + {b_m} + {b_{m + 1}} = n} {\frac{{n!}}{{{b_1}!{b_2}! \ldots {b_m}!{b_{m + 1}}!}}} \)

\( = \begin{array}{*{20}{c}}\sum \\{{b_1} + {b_2} +  \cdots  + {b_m} + {b_{m +1}}=n}\end{array}\left( {\begin{array}{*{20}{c}}n\\{{b_1},{b_2}, \ldots ,{b_m},{b_{m + 1}}}\end{array}} \right)\)

Now, we can write as

\(\prod\limits_{j = 1}^{m – 1} {x_j^{{b_j}}}  \cdot x_m^{{b_m}}x_{m + 1}^{{b_{m + 1}}} = x_1^{{b_1}} \cdot x_2^{{b_2}} \cdot x_3^{{b_3}} \cdots x_{m – 1}^{{b_{m – 1}}} \cdot x_m^{{b_m}} \cdot x_{m + 1}^{{b_{m + 1}}}\)

\( = \mathop \prod \limits_{j = 1}^{m + 1} x_j^{{b_j}}\)

Now, substituting all the values in the expansion, we get

\(\left(x_{1}+x_{2}+x_{3}+\cdots+x_{m-1}+\left(x_{m}+x_{m+1}\right)\right)^{n}\)

\( = \sum\limits_{{b_1} + {b_2} +  \cdots  + {b_{m- 1}}+M=n}{\left( {\begin{array}{*{20}{c}}n\\{{b_1},{b_2}, \ldots ,{b_{m – 1}},M}\end{array}} \right)} \sum\limits_{{b_m} + {b_{m +1}}=M}{\left( {\begin{array}{*{20}{c}}M\\{{b_m},{b_{m + 1}}}\end{array}} \right)}\)

\( \cdot \prod\limits_{j = 1}^{m – 1} {x_j^{{b_j}}}  \cdot x_m^{{b_m}}x_{m + 1}^{{b_{m + 1}}}\)

\(\left(x_{1}+x_{2}+x_{3}+\cdots+x_{m}+x_{m+1}\right)^{n}=\) \(\begin{array}{*{20}{c}}\sum \\{{b_1} + {b_2} +  \cdots  + {b_m} + {b_{m +1}}=n}\end{array}\left( {\begin{array}{*{20}{c}}n\\{{b_1},{b_2}, \ldots ,{b_m},{b_{m + 1}}}\end{array}} \right)\prod\limits_{j = 1}^{m + 1} {x_j^{{b_j}}} \)

Now, if we substitute \(k=m+1\) in \(RHS\) of the multinomial theorem, then

\(RHS = \sum\limits_{{b_1} + {b_2} +  \cdots  + {b_k} = n} {\left( {\begin{array}{*{20}{c}}n\\{{b_1},{b_2}, \ldots ,{b_k}}\end{array}} \right)} \prod\limits_{j = 1}^k {x_j^{{b_j}}} \)

\(\therefore R H S=\sum\limits_{{b_1}+{b_2}+\cdots+{b_m}+{b_{m+1}}=n}{\left( {\begin{array}{*{20}{c}}n\\{{b_1},{b_2}, \ldots ,{b_m},{b_{m + 1}}}\end{array}} \right)} \prod\limits_{j = 1}^{m + 1} {x_j^{{b_j}}} \)

Thus, \(LHS=RHS\)

Hence, we can see that the multinomial theorem is true for \(k=m+1\) if valid for \(k= m\), and since \(k=1\) is true.

Thus, we can say that the multinomial theorem holds for all values \(k\), such that \(k\) is a natural number.

Hence, the multinomial theorem is proved.

Number of Distinct Terms

The number of terms in the expansion \(\left(x_{1}+x_{2}+x_{3}+\ldots+x_{k}\right)^{n}\) is equal to the number of non-negative integer solutions of the equation, \(b_{1}+b_{2}+ \ldots+b_{k}=n\) because each solution of this equation gives a term in the expansion.

The number of such solutions is given by

\({}^{n+k-1} C_{k-1}\)

Greatest Coefficient in a Multinomial Expansion

Greatest coefficient in the expansion of \(\left(x_{1}+x_{2}+x_{3}+\cdots+x_{k}\right)^{n}\) is

\(\frac{n !}{(q !)^{k-r}((q+1) !)^{r}}\)

Where,

\(q\) is the quotient

\(r\) is the remainder when \(n\) is divided by \(k\).

General Term in Multinomial Expansion

The general term in the multinomial expansion \(\left(x_{1}+x_{2}+x_{3}+\cdots+x_{k}\right)^{n}\) is

\(\frac{n !}{b_{1} ! b_{2} ! b_{3} ! \ldots . b_{k} !} \cdot x_{1}^{b_{1}} x_{2}^{b_{2}} x_{3}^{b_{3}} \ldots \ldots x_{k}^{b_{k}}\)

where \(b_{1}+b_{2}+ \ldots+b_{k}=n\)

Solved Examples – Multinomial Theorem

Q.1. Find the total terms in the expansion of \(\left(x_{1}+x_{2}+x_{3}\right)^{4}\)
Ans: Given that the expansion is \(\left(x_{1}+x_{2}+x_{3}\right)^{4}\)
The total number of terms \({ }^{4+3-1} C_{3-1}={ }^{6} C_{2}\)
\(=\frac{6 !}{2 !(6-2) !}\)
\(=\frac{6 !}{2 ! 4 !}\)
\(=\frac{6 \times 5 \times 4 !}{2 ! 4 !}\)
\(=15\)
Hence, the number of terms of the given expansion \(\left(x_{1}+x_{2}+x_{3}\right)^{4}\) is \(15\).

Q.2. Find the coefficient of \(x^{2} y^{3} z^{4} w\) in the expansion of \((x-y-z+w)^{10}\)
Ans: Given that, the expansion is \((x-y-z+w)^{10}\)
Therefore, \((x-y-z+w)^{10}=\sum_{p+q+r+s=10} \frac{n !}{p ! q ! r ! s !}(x)^{p}(-y)^{q}(-z)^{r}(w)^{s}\)
We have to find the coefficient of \(x^{2} y^{3} z^{4} w\), which implies that \(p=2, q=3\), \(r=4, s=1\)
Hence, the coefficient of \(x^{2} y^{3} z^{4} w\) is \(\frac{10 !}{2 ! 3 ! 4 ! 1 !}(-1)^{3}(-1)^{4}=-12600\)

Q.3. Determine the coefficient of \(x^{5}\) in the expansion \(\left(2-x+3 x^{2}\right)^{6}\).
Ans: Given that the expansion is \(\left(2-x+3 x^{2}\right)^{6}\)
The general term in the expansion of \(\left(2-x+3 x^{2}\right)^{6}=\frac{6 !}{r ! s ! t !} 2^{r}(-x)^{s}\left(3 x^{2}\right)^{t}\), where \(r+s+t=6\)
\(=\frac{6 !}{r ! s ! t !} 2^{r} \times(-1)^{s} \times(3)^{t} \times x^{s+2 t}\)
For the coefficient of \(x^{5}\), we must have \(s+2 t=5\), but we have \(r+s+t=6\)
\(\therefore s=5-2 t\) and \(r=1+t\), where \(0 \leq r, s, t \leq 6\)
Now, \(t=0 \Rightarrow r=1, s=5\)
\(t=1 \Rightarrow r=2, s=3\)
\(t=2 \Rightarrow r=3, s=1\)
Thus, the three containing \(x^{5}\) and coefficient of \(x^{5}=\frac{6 !}{1 ! 5 ! 0 !} \times 2^{1} \times(-1)^{5} \times 3^{0}+\frac{6 !}{2 ! 3 ! 1 !} \times 2^{2} \times(-1)^{3} \times 3^{1}+\frac{6 !}{3 ! 1 ! 2 !} \times 2^{3} \times(-1)^{1} \times 3^{2}\)
\(=-12-720-4320\)
\(=-5052\)
Hence, the coefficient of \(x^{5}\) of expansion \(\left(2-x+3 x^{2}\right)^{6}\) is \(-5052\).

Q.4. Determine the total number of terms in the expansion of \((1+x+y)^{10}\)
Ans: Given that the expansion is \((1+x+y)^{10}\)
So, the total number of terms \(={}^{10+3-1} C_{3-1}\)
\(={}^{12} C_{2}\)
\(=\frac{12 !}{2 !(12-2) !}\)
\(=\frac{12 !}{2 ! 10 !}\)
\(=\frac{12 \times 11 \times 10 !}{2 ! 10 !}\)
\(=66\)
Hence, the number of terms of the given expansion \((1+x+y)^{10}\) is \(66\).

Q.5. Find the coefficient of \(x^{2} y^{3}\) in the expansion of \((1+x+y)^{10}\).
Ans: Given that the expansion is \((1+x+y)^{10}\)
Therefore, \((1+x+y)^{10}=\sum_{p+q+r=10} \frac{n !}{p ! q ! r !}(1)^{p}(x)^{q}(y)^{r}\)
We have to find the coefficient of \(x^{2} y^{3}\), which implies that, \(p=5, q=2, r=3\)
Hence, the coefficient of \(x^{2} y^{3}=\frac{10 !}{2 ! \times 3 ! \times 5 !}\)
\(=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{2 ! \times 3 ! \times 5 !}\)
\(=2520\)

Summary

The binomial theorem states the principle of expanding the algebraic expression \((x+y)^{n}\), and expresses it as a sum of the terms involving individual exponents of variables \(x\) and \(y\). The multinomial theorem is generally used to expand the algebraic expressions, which have more than two terms with has higher exponents. The multinomial theorem generalises the binomial theorem to include polynomials with any number of terms. We learned about the proof of the multinomial theorem using the principle of mathematical induction. Later, the multinomial coefficient, general term, the number of terms, and the greatest coefficient were explained.

Frequently Asked Questions (FAQs)

Q.1. What is multinomial theorem in binomial?
Ans: The multinomial theorem, in algebra, a generalisation of the binomial theorem to more than two variables.
\({\left( {{x_1} + {x_2} +  \cdots  + {x_k}} \right)^n} = \sum\limits_{{n_1},{n_2}, \cdots ,{n_k} \ge 0} {\frac{{n!}}{{{n_1}!{n_2}! \cdots {n_k}!}}} x_1^{{n_1}}x_2^{{n_2}} \cdots x_k^{{n_k}}\)
where \(n_{1}+n_{2}+\cdots+n_{k}=n\).

Q.2. What is multinomial and example?
Ans: An algebraic expression that has two or more terms is called a multinomial.
Example: \(9 x^{3}+2 x^{2}+4\)

Q.3. What is the multinomial theorem used for?
Ans: The multinomial theorem is used to expand the power of a sum of two or more terms. It is mainly used to generalise the binomial theorem to polynomials with any number of terms.

Q.4. How many different terms are there in the multinomial expansion?
Ans: The total number of terms in the expansion \(\left(x_{1}+x_{2}+\cdots+x_{k}\right)^{n}\) is equal to the number of non-negative integer solutions that is given by \({}^{n + k – 1}{C_{k – 1}}\)

Q.5. What is the general term in the multinomial expansion?
Ans: The general term of the expansion \(\left(x_{1}+x_{2}+x_{3}+\cdots+x_{k}\right)^{n}\) is
\(\frac{n !}{b_{1} ! b_{2} ! b_{3} ! \ldots . b_{k} !} \cdot x_{1}^{b_{1}} x_{2}^{b_{2}} x_{3}^{b_{3}} \ldots \ldots x_{k}^{b_{k}}\)
Where, \(b_{1}+b_{2}+\cdots \ldots+b_{k}=n\)

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