Question

Find the following multinomials: (a) The coefficient of \(x^{2} y^{3} z^{5}\) in the expansion of \((x+y+z)^{10}\) (b) The coefficents of \(x^{3} y^{4} z^{2}\) and \(x^{3} y^{3} z^{3}\) in the expansion of \((3 x+2 y-z)^{9}\)

Short answer

(a) 252; (b.i) 725760; (b.ii) -233280.

Step-by-step solution

Identify the General Formula

To find the coefficient of any term in the expansion of \((a + b + c)^n\), use the multinomial theorem. The appropriate term for \(x^p y^q z^r\) is given by \[ \frac{n!}{p!q!r!}a^p b^q c^r \] for \(p + q + r = n.\) Here, we'll keep \(a = x, b = y, c = z.\)

Calculate the Coefficient for Part (a)

We want the coefficient of \(x^2 y^3 z^5\) in the expansion of \((x + y + z)^{10}\). Here, \(n = 10, p = 2, q = 3, r = 5.\) \[ \frac{10!}{2!3!5!} = \frac{3628800}{2 \times 6 \times 120} = 252 \] Thus, the coefficient is 252.

Set Up the Expression for Part (b.i) and Use the Formula

For \(x^3 y^4 z^2\) in the expansion of \((3x + 2y - z)^9\): Let \(a = 3x, b = 2y, c = -z\). Here \(n = 9, p = 3, q = 4, r = 2.\)\[ \frac{9!}{3!4!2!} (3)^3 (2)^4 (-1)^2 = \frac{362880}{6 \times 24 \times 2} \times 27 \times 16 \times 1 = 1680 \times 27 \times 16 = 725760 \] Thus, the coefficient is 725760.

Calculate Coefficient for Part (b.ii)

For the term \(x^{3} y^{3} z^{3}\) in the expansion of \((3x + 2y - z)^9\): With \(p = 3, q = 3, r = 3\), use\[ \frac{9!}{3!3!3!} (3)^3 (2)^3 (-1)^3 = \frac{362880}{6 \times 6 \times 6} \times 27 \times 8 \times (-1) = -1080 \times 27 \times 8 = -233280 \] Thus, the coefficient is -233280.

Key concepts

Multinomial Theorem

The multinomial theorem is an extension of the binomial theorem. It helps in expanding any expression raised to a power. If you have an expression like \( (x_1 + x_2 + ... + x_m)^n \), it tells us how to find the expanded form.
It involves finding the coefficients for each term in the expansion. These coefficients are known as multinomial coefficients.
If you're looking for the coefficient of a specific term \( x_1^{k_1} x_2^{k_2} ... x_m^{k_m} \) in the expansion, you use the formula:\[\frac{n!}{k_1!k_2!...k_m!}x_1^{k_1}x_2^{k_2}...x_m^{k_m}\]where \( k_1 + k_2 + ... + k_m = n \).

• The expression \( n! \) denotes "n factorial," which is the product of all positive integers up to \( n \).
• The terms \( x_1^{k_1}, x_2^{k_2}, ... \) represent the variables raised to specific powers.

This theorem is fundamental in combinatorics since it provides us with a systematic way of determining the number of ways to arrange groups of items.
It is highly useful in real-world scenarios, such as calculating probabilities in probabilistic models, analyzing algorithms, and more.

Polynomial Expansion

Polynomial expansion involves writing a power of a polynomial as a sum of terms. Each term consists of a product of the polynomial's variables raised to different powers. This concept relates closely to the multinomial theorem. The expansion of a multinomial function like \( (a + b + c)^n \) can be obtained using the theorem's formula.
In this process:

• Each expansion term is formed as a combination of different powers of each variable.
• The overall sum of these powers in a term equals the polynomial's exponent.
• Each term's coefficient is determined using the multinomial coefficients.

For instance, expanding \( (x + y + z)^{10} \) using polynomial expansion and multinomial theorem, allows us to find specific terms such as \( x^2 y^3 z^5 \).
This systematic method helps in simplifying, solving polynomial equations, and is key in various formulas in algebra.

Combinatorics

Combinatorics is a field of mathematics concerning the counting, arrangement, and combination of objects. It provides the mathematical tools to answer questions such as "How many ways can we select objects from a group?" or "In how many ways can we organize these objects?"
Applications of combinatorics are extensive, ranging from game theory to coding theory.

• Multinomial coefficients, used in the multinomial theorem, are key components of combinatorics. They count how many ways you can distribute \( n \) items into different groups.
• In combinatorics, factorials \( n! \) often find use in counting permutations, which are different orderings of objects.
• Combinatorics also deals with combinations, which involve selecting objects without regard to order, unlike permutations.

Understanding these mathematical principles can significantly enhance problem-solving techniques. Whether considering simple problems like arranging books on a shelf or more complex ones like calculating possibilities in the game of poker, combinatorics offers a robust framework.