Multinomial Theorem formula expansion method and solved examples
The multinomial theorem is the extended form of the binomial theorem. It describes the result of expanding the power of a multinomial. The multinomial theorem describes how we can expand the power of a sum that consists of more than two terms. It is a generalization of the binomial theorem to a polynomial with any number of terms. It expresses a power \[(x_{1} + x_{2} +...+x_{n})^{n}\] as a weighted sum of monomials of for \[\mathrm{x_{1}^{b^{1}}x_{2}^{b^{2}}x_{n}^{b^{n}}}\] where the weights are given by generalizations of binomial coefficient and multinomial coefficient.
In this article, we will learn about no of terms in multinomial expansion, multinomial coefficient proof and the relation between binomial and multinomial theorems.
Consider (a+b+c)4. Using brute force way of expanding this we can write it as (a + b + c) (a + b + c) (a + b + c) (a + b + c),then we will apply the distributive law and after that simplify it by collecting like terms. After distributing, but before collecting like terms, there will be 81 terms. (This is because every term in the first brackets has to be multiplied by every term in the second brackets, which will give 9 terms. Each of these terms 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, which will give 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 the final result is shown below. We need to consider what distinct terms will occur and how many ways there are of getting each of them:
(a+b+c)4=(a+b+c) (a+b+c) (a+b+c) (a+b+c)
Here we will use multinomial theorem expansion
a4+b4+c4+4ab3+4ac3+4a3b+4a3c+4bc3+4b3c+12abc2+12ab2c+12a2bc+6a2b2+6a2c2+6b2c2
Multinomial coefficient formula
Given below is the multinomial coefficient formula
\[(\frac{n}{k_{1},k_{2}.....k_{r-1}}) = (\frac{n!}{k_{1}!k_{2}!....k_{r-1}!k_{r}})\]
Multinomial Theorem Statement
For a positive integer m and a non-negative integer n, the sum of m terms raised to the power n is expended as
\[\mathrm{(x_{1} + x_{2} +......x_{m})^{n} = \frac{\sum}{k_{1} + k_{2} +......k_{m} = n}\left ( \frac{n}{{k_{1},k_{2}.....k_{m}}} \right )\prod_{t=1}^{m}x_k^kt}\]
Where \[(\frac{n}{k_{1},k_{2}.....k_{m}}) = (\frac{n!}{k_{1}!k_{2}!....k_{m}!})\] is the multinomial coefficient.
The number of terms of this sum is given by a stars and bars argument: \[\frac{n+k-1}{n}\]
Multinomial Theorem Proof
Multinomial theorem proof can be done by two types.
An algebraic proof by induction
A combinatorial proof by counting.
Proof of multinomial theorem is given below
Consider a positive integer k and a non-negative integer n
\[\mathrm{(x_{1} + x_{2} + x_{3} +...x_{k-1} + x_{k})^{n} = \frac{\sum}{b_{1} + b_{2} + b_{3}...+ b_{k-1}+ b_{k}= n}\left ( \frac{n}{{b_{1},b_{2},b_{3}.....b_{k-1},b_{k}}} \right ) \prod_{j=1}^{k}x_j^bj}\]
When k=1 the result is true and when k=2 the result is the binomial theorem. Assume that \[k = \underline{>}3\] and that the result is true for k=p and k=p+1
\[\mathrm{(x_{1} + x_{2} + x_{3} +...x_{(p-1)} + x_{p})^{n}= (x_{1} + x_{2} + x_{3} +...x_{(p-1)} + (x_{p} + x_{(p+1)}))^{n}}\]
Treating xp+xp+1as a single term and using the induction hypothesis:
\[\mathrm{\frac{\sum}{b_{1} + b_{2} + b_{3}...+ b_{p-1}+ B= n}\left ( \frac{n}{{b_{1},b_{2},b_{3}.....b_{p-1}, B}} \right ) . (x_{p} + x_{p+1})^{B}.\prod_{j=1}^{p-1}x_j^bj}\]
By the Binomial Theorem, this becomes:
\[\mathrm{\frac{\sum}{b_{1} + b_{2} + b_{3}...+ b_{p-1}+ B= n}(\frac{n}{{b_{1},b_{2},b_{3}.....b_{p-1}, B}}) (\prod_{j=1}^{p-1}x_j^bj)\sum_{b_{p}+b_{p+1}=B} (\frac{B}{b_{p}}). x_p^bp \ x_{p+1}^{bp+1}}\]
Since \[\mathrm{\left ( \frac{n}{{b_{1},b_{2},b_{3}.....b_{p}, B}} \right )\left ( \frac{B}{b_{p}} \right )\: = \left ( \frac{n}{{b_{1},b_{2},b_{3}.....b_{p}, b_{p+1}}} \right )}\] this can be rewritten as
\[\mathrm{\frac{\sum}{b_{1} + b_{2} + b_{3}...+ b_{p+1}= n}\left ( \frac{n}{{b_{1},b_{2},b_{3}.....b_{p+1}}} \right )\prod_{j=1}^{k}x_j^bj}\]
Number of terms in a multinomial expansion
Consider the multinomial expression \[(x_{1} + x_{2} + x_{3} +x_{4} + x_{5} +... + x_{m})^{n}\]
The total no of terms in multinomial expansion of the above expression will be \[^{n+m-1}C_{n}\].
Greatest coefficient in multinomial expansion
Consider the multinomial expression \[(x_{1} + x_{2} + x_{3} +x_{4} + x_{5} +... + x_{m})^{n}\]
Let q be the quotient and r be the remainder when nis divided by m. Then, the greatest coefficient in the multinomial expression given above is \[\frac{n!}{(q!)^{m-r}((q+1)!)^{r}}\]
Sum of coefficients in multinomial expansion
The sum of coefficients in multinomial expression can be obtained easily by putting the value of all the variables as 1 in the multinomial expression.
For Example: Consider the multinomial expression \[(x + 2y + z)^{n} \]
Putting x = 1, y = 1 and z =1 in this multinomial expression becomes (4)n
Hence, the sum of coefficients in the above multinomial expression is (4)n
Multinomial theorem in permutation and combination
The multinomial coefficient can be used in permutations and combinations to find the number of distinguishable permutations of n objects when \[n = n_{1} + n_{2} +... + n_{k}\] and we have n1 items of kind 1, n2 items of type 2, and nk items of type k for every k.
Conclusion:
From the above article, we get to know about the multinomial theorem and its proof. The multinomial theorem gives the sum of multinomial coefficients multiplied by variables. In other words, we can say it is used to represent an expanded series where each term in it has its own associated multinomial coefficient. We know how to calculate the total number of terms in multinomial expansion and various applications of it.
FAQs on Multinomial Theorem Explained with Formula and Proof
1. What is the Multinomial Theorem?
The Multinomial Theorem gives the expansion of a power of a sum of more than two terms. It states that for a positive integer n, (x₁ + x₂ + ... + x_k)^n = ∑ [n!/(n₁!n₂!...n_k!)] x₁^{n₁} x₂^{n₂} ... x_k^{n_k}, where n₁ + n₂ + ... + n_k = n.
- It is a generalization of the Binomial Theorem.
- The coefficients are called multinomial coefficients.
- Each term corresponds to a specific distribution of n among the variables.
2. What is the formula for the Multinomial Theorem?
The formula for the Multinomial Theorem is (x₁ + x₂ + ... + x_k)^n = ∑ [n!/(n₁!n₂!...n_k!)] x₁^{n₁} x₂^{n₂} ... x_k^{n_k}, where n₁ + n₂ + ... + n_k = n.
- n! is the factorial of n.
- The denominator contains factorials of the exponents.
- The sum runs over all non-negative integer solutions of n₁ + n₂ + ... + n_k = n.
3. How do you expand (x + y + z)^n using the Multinomial Theorem?
To expand (x + y + z)^n, apply the formula n!/(a!b!c!) x^a y^b z^c for all a + b + c = n.
- Step 1: List all non-negative integer triples (a, b, c) such that a + b + c = n.
- Step 2: Compute the coefficient n!/(a!b!c!).
- Step 3: Write each term as x^a y^b z^c.
4. What is a multinomial coefficient?
A multinomial coefficient is the number n!/(n₁!n₂!...n_k!) that appears in the expansion of (x₁ + x₂ + ... + x_k)^n.
- It counts the number of ways to partition n identical items into groups of sizes n₁, n₂, ..., n_k.
- It is a generalization of the binomial coefficient nCr.
- When k = 2, it reduces to n!/(r!(n−r)!).
5. How is the Multinomial Theorem different from the Binomial Theorem?
The Multinomial Theorem expands powers of sums with more than two terms, while the Binomial Theorem applies only to two terms.
- Binomial: (a + b)^n.
- Multinomial: (a + b + c + ...)^n.
- The binomial coefficient nCr is a special case of the multinomial coefficient.
6. Can you give an example of the Multinomial Theorem?
An example of the Multinomial Theorem is the expansion of (x + y + z)^2 = x² + y² + z² + 2xy + 2xz + 2yz.
- Possible exponent combinations: (2,0,0), (1,1,0), etc.
- Coefficient for (1,1,0) is 2!/(1!1!0!) = 2.
- Each term's exponents add up to 2.
7. How many terms are there in a multinomial expansion?
The number of terms in (x₁ + x₂ + ... + x_k)^n is (n + k − 1)! / (n!(k − 1)!).
- This counts the number of non-negative integer solutions to n₁ + n₂ + ... + n_k = n.
- For (x + y + z)², the number of terms is (2+3−1)!/(2!2!) = 6.
- This formula is based on combinations with repetition.
8. What are the conditions for using the Multinomial Theorem?
The Multinomial Theorem applies when expanding (x₁ + x₂ + ... + x_k)^n where n is a non-negative integer.
- Each exponent n₁, n₂, ..., n_k must be a non-negative integer.
- The sum of the exponents must equal n.
- The formula works for algebraic variables and numerical terms.
9. How do you find a specific term in a multinomial expansion?
To find a specific term in (x₁ + x₂ + ... + x_k)^n, use n!/(n₁!n₂!...n_k!) x₁^{n₁} x₂^{n₂} ... x_k^{n_k} where the exponents match the required powers.
- Step 1: Identify the desired exponents.
- Step 2: Ensure their sum equals n.
- Step 3: Substitute into the multinomial coefficient formula.
10. What are common mistakes when using the Multinomial Theorem?
Common mistakes in the Multinomial Theorem include incorrect factorial calculations and forgetting that exponents must sum to n.
- Not ensuring n₁ + n₂ + ... + n_k = n.
- Incorrectly computing factorial values.
- Missing terms or repeating equivalent exponent combinations.
