Multinomial Expansion

Many of you may have heard about binomial expansion (i.e., expanding a polynomial of two variables to their nth power). But what happens if we have to calculate ? If we start using the method of binomial expansion, it will take an extremely long time to reach the full polynomial. Hence, I am introducing a more general approach that includes binomial expansion. It is known as multinomial expansion. By this method, we can generate polynomials like the ones given in the above example.

Let me demonstrate a simple calculation by computing by brute-force calculation:

$$ \text{Example 1:} \space (x+y+z)^3 \newline = (x+y+z) \times (x+y+z) \times (x+y+z) \newline = ((x+y+z) \times (x+y+z)) \times (x+y+z) $$
$$ \text{We will solve the big bracket and then multiply the two brackets.} \newline = ({x^2+y^2+z^2+(x \cdot y)+(x \cdot z)+(y \cdot x)+(y \cdot z)+(z \cdot x)+(z \cdot y)}) $$
$$ \text{But, from multiplication, $x \cdot y$ = $y \cdot x$; therefore $(y \cdot x)+(x \cdot y) = 2 \times (x \cdot y)$} \newline \text{Therefore, $({x^2}+{y^2}+{z^2}+2 \times (xy+xz+yz)$} $$
$$ \text{Now to obtain our full polynomial, we will have to multiply the above result with $(x+y+z)$.} \newline = ({x^2}+{y^2}+{z^2}+2 \times (xy+xz+yz)) \times (x+y+z) $$
$$ = x^3+(x \cdot y^2)+(x \cdot z^2)+(2 \times x^2 \cdot y)+(2 \times x^2 \cdot z)+(2 \times x y z) \newline +(y \cdot x^2)+(y^3)+(y \cdot z^2)+(2 \times x \cdot y^2)+(2 \times x y z)+(2 \times y^2 \cdot z) \newline +(z \cdot x^2)+(z \cdot y^2)+(z^3)+(2 \times x y z)+(2 \times x \cdot z^2)+(2 \times y \cdot z^2) $$
$$ = x^3+y^3+z^3+3 \times ({x^2} y+{y^2} x+x z^2+{x^2} z+{y^2} z+{z^2} y)+(6 \times x y z)$$

$$ \text{(Note: In the expression, the dot between two letters is used as multiplication symbol,} \newline \text{like found in most mathematical books.)} $$

As you can see, the polynomial can be calculated, but it becomes a very tedious process as we move higher with more variables and higher powers, and there are high chances that we will make errors in counting. Hence, there is a simple but effective formula to calculate every term in its general form, which is shown as follows:

$$ ({x_1}+{x_2}+{x_3}+{x_4}+….{x_k})^n = \newline \sum_{{n_1}+{n_2}+…{n_k}=n}{(\frac{n!}{{n_1!} \cdot {n_2!} \cdot {n_3!}… \cdot {n_k!}}) \times ({x_1^{n_1}} \cdot {x_2^{n_2}} \cdot {x_3^{n_3}} … \cdot {x_k^{n_k}})} $$

We will now see an example of the following multinomial expansion.

$$ (x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7)^7 $$

Now, the above expression is a summation of 7 different terms raised to the power of 7. Like I have shown in the above example, if it were just a square or cube of the expression within the parenthesis, it would be an easy task to calculate the polynomial. But to do the same thing 7 times is a very challenging and time-consuming task if done by brute-force method. Hence, we will follow the procedure like this:

1. Analyze the powers of the terms.
2. Calculate the multinomial coefficient using the formula.
3. Apply the same for all combinations of powers whose sum should exactly be equal to 7 (for this given example).
4. Write the final expression in summation notation of the terms, their powers, and the multinomial coefficients.

Possible power combinations:

1. 7 (each term raised to power 7) ()
2. 6 + 1 ()
3. 5 + 2 ()
4. 5 + 1 + 1 (The product looks like )
5. 4 + 3 ( (Where, i and j can be 1 to 7, but )
6. 4 + 2 + 1 (The product looks like )
7. 4 + 1 + 1 + 1 (The product looks like )
8. 3 + 3 + 1 (The product looks like )
9. 3 + 2 + 2 (The product looks like )
10. 3 + 2 + 1 + 1 (The product looks like )
11. 3 + 1 + 1 + 1 + 1 (The product looks like )
12. 2 + 2 + 2 + 1 (The product looks like )
13. 2 + 2 + 1 + 1 + 1 (The product looks like )
14. 2 + 1 + 1 + 1 + 1 + 1 (The product looks like )
15. 1 + 1 + 1 + 1 + 1 + 1 + 1 ()

Multinomial Coefficients for the power combinations: (Note: the sum of numbers in the denominator will be equal to 7 at all times)

$$1. \space \frac{7!}{7!} = 1 $$

$$2.\space \frac{7!}{6!\times1} = 7 $$

$$3. \space \frac{7!}{5!\times 2!} = 21 $$

$$4. \space \frac{7!}{5!\times1\times1} = 42 $$

$$5. \space \frac{7!}{4!\times3!} = 35 $$

$$6. \space \frac{7!}{4!\times2!\times1} = 105 $$

$$7. \space \frac{7!}{4!\times1\times1\times1} = 210 $$

$$8. \space \frac{7!}{3!\times3!\times1} = 140 $$

$$9. \space \frac{7!}{3!\times2!\times2!} = 210 $$

$$10. \space \frac{7!}{3!\times2!\times1\times1} = 420 $$

$$11. \space \frac{7!}{3!\times1\times1\times1\times1} = 840 $$

$$12. \space \frac{7!}{2!\times2!\times2!\times1} = 630 $$

$$13. \space \frac{7!}{2!\times2!\times1\times1\times1} = 1260 $$

$$14. \space \frac{7!}{2!\times1\times1\times1\times1\times1} = 2520 $$

$$15. \space \frac{7!}{1\times1\times1\times1\times1\times1\times1} = 5040 $$

After having calculated the power combinations and the multinomial coefficients, the only task left to do is write the summation of all the terms to write the polynomial. (Remember: The terms and their powers are written to show how all the terms in that particular category will be in the final polynomial.) Therefore, the polynomial is as follows:

$$ ({x_1}+{x_2}+{x_3}+{x_4}+{x_5}+{x_6}+{x_7})^7 = \newline \sum {x_i}^7 + 7\times \sum{{x_i}^6\cdot{x_j}} \newline + 21\times \sum{{x_i}^5\cdot{x_j}^2} + 42\times \sum{{x_i}^5\cdot{x_j}\cdot{x_k}} \newline + 35\times \sum{{x_i}^4\cdot{x_j}^3} + 105\times \sum{{x_i}^4\cdot{x_j}^2\cdot{x_k}} \newline + 210\times \sum{{x_i}^4\cdot{x_j}\cdot{x_k}\cdot{x_l}} + 140\times \sum{{x_i}^3\cdot{x_j}^3\cdot{x_k}} \newline + 210\times \sum{{x_i}^3\cdot{x_j}^2\cdot{x_k}^2} + 420\times \sum{{x_i}^3\cdot{x_j}^2\cdot{x_k}\cdot{x_l}} \newline + 840\times \sum{{x_i}^3\cdot{x_j}\cdot{x_k}\cdot{x_l}\cdot{x_m}} + 630\times \sum{{x_i}^2\cdot{x_j}^2\cdot{x_k}^2\cdot{x_l}} \newline + 1260\times \sum{{x_i}^2\cdot{x_j}^2\cdot{x_k}\cdot{x_l}\cdot{x_m}} + 2520\times \sum{{x_i}^2\cdot{x_j}\cdot{x_k}\cdot{x_l}\cdot{x_m}\cdot{x_n}} \newline + 5040\times {x_1\cdot{x_2}\cdot{x_3}\cdot{x_4}\cdot{x_5}\cdot{x_6}\cdot{x_7}} $$

In summary, the multinomial theorem provides a powerful generalization of the binomial theorem, enabling us to expand expressions of the form $(x_1+x_2+\cdots +x_m{)}^n$ in a systematic way. Each term in the expansion corresponds to a unique combination of non‑negative integer exponents summing to $n$, with the coefficient given by the multinomial coefficient $\frac{n!}{k_1!k_2!\dots k_m!}$.

Understanding the structure of these terms—grouped by the pattern of positive exponents (i.e., integer partitions of $n$ into at most $m$ parts)—reveals not only how many terms of each type appear but also their exact coefficients. This framework has applications across combinatorics, probability (e.g., multinomial distributions), algebra, and number theory.

By mastering the multinomial theorem, one gains a versatile tool for handling polynomial expansions with multiple variables, laying the groundwork for more advanced topics such as generating functions, symmetric polynomials, and combinatorial identities. I’ll end the article here. Until next time, stay curious and keep learning!