How to Use the Multinomial Theorem to Expand Expressions
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 Formulas and Examples
1. What is a multinomial expression?
A multinomial expression is simply an algebraic expression that contains more than two terms. For example, while (x + y) is a binomial, an expression like (x + y + z) is a multinomial. The Multinomial Theorem is the tool used to expand these longer expressions.
2. What does the Multinomial Theorem explain?
The Multinomial Theorem provides a direct formula to expand a multinomial raised to a power, such as (x₁ + x₂ + ... + xₖ)ⁿ. It tells you exactly how to find all the resulting terms and their specific coefficients without having to multiply the expression manually.
3. What is the general formula used in the Multinomial Theorem?
The expansion of (x₁ + x₂ + ... + xₖ)ⁿ is the sum of terms that look like this:
[n! / (n₁! * n₂! * ... * nₖ!)] * (x₁ⁿ¹ * x₂ⁿ² * ... * xₖⁿᵏ)
For each term, the sum of the powers (n₁ + n₂ + ... + nₖ) must equal 'n'. The first part of this formula is known as the multinomial coefficient.
4. How do you find the coefficient of a specific term in a multinomial expansion?
To find the coefficient of a specific term, like x₁ᵃx₂ᵇ...xₖᶜ in the expansion of (x₁ + ... + xₖ)ⁿ, you just need to use the multinomial coefficient formula: n! / (a! * b! * ... * c!). Remember that the sum of the powers (a + b + ... + c) must equal the total power, n.
5. Could you show a simple example of using the Multinomial Theorem?
Certainly. Let's find the term containing x¹y²z¹ in the expansion of (x + y + z)⁴. Here, n=4, and the powers are 1, 2, and 1 (since 1+2+1=4).
The coefficient is calculated as: 4! / (1! * 2! * 1!) = 24 / (1 * 2 * 1) = 12.
Therefore, the complete term is 12x¹y²z¹.
6. What is the main difference between the Binomial and Multinomial Theorems?
The key difference is the number of terms in the base expression.
- The Binomial Theorem is specifically for expanding expressions with two terms, like (a + b)ⁿ.
- The Multinomial Theorem is the general version that works for any expression with two or more terms, like (a + b + c)ⁿ.
7. How is the Multinomial Theorem related to permutations and combinations?
The connection is in the coefficient. The multinomial coefficient formula, n! / (n₁!n₂!...nₖ!), is the same one used in combinatorics to find the number of ways to arrange 'n' objects into 'k' distinct groups. This is why it's used to count how many times each specific term appears in the expansion.
8. Is the Multinomial Theorem covered in the Class 11 CBSE syllabus?
The Binomial Theorem is a core part of the Class 11 Maths syllabus as per NCERT. The Multinomial Theorem, however, is generally treated as an extension of this topic. While it might not be a focus for school board exams, it is a very important concept for competitive exams like JEE Main and Advanced.
9. Why is the Multinomial Theorem useful in a field like probability?
In probability, the theorem is used to solve problems for experiments with more than two possible outcomes, known as a multinomial distribution. For example, if you roll a die many times, the Multinomial Theorem can help you calculate the exact probability of getting a specific count of 1s, 2s, 3s, and so on.
