# What is the coefficient of the x^2 y^4 term in the expansion of (x + y)^6?

## Question:

What is the coefficient of the {eq}x^2 y^4 {/eq} term in the expansion of {eq}(x + y)^6 {/eq}?

## Binomial Theorem:

For a non-negative integer 'n', {eq}(x + y)^n {/eq} is solved using the binomial expansion. The general formula for binomial expansion is:

{eq}(x + y)^n = \binom{n}{0}x^n y^0 + \binom{n}{1}x^{n-1} y^1 + \binom{n}{2}x^{n-2} y^2 + \ .......\ + \binom{n}{n-1}x^{1} y^{n-1} + \binom{n}{n}x^0 y^n {/eq}

Where, {eq}\binom{n}{r} = \dfrac{n!}{r! (n-r)!} {/eq} is a positive integer known as the binomial coefficient.

## Answer and Explanation:

Using the formula of binomial expansion, we can solve as:

{eq}(x+y)^6 = \binom{6}{0} x^6 y^0 + \binom{6}{1} x^5 y^1 + \binom{6}{2} x^4 y^2 + \binom{6}{3} x^3 y^3+ \binom{6}{4} x^2 y^4 + \binom{6}{5} x^1 y^5 + \binom{6}{6} x^0 y^6 \\ {/eq}

We can expand this fully and solve for it, but for finding the coefficients of a particular term we use following procedure.

**Step one:** write the general form of the expansion.

{eq}(x+y)^6 = \sum_{r=0}^{6} \binom{6}{r} x^{6-r} y^r {/eq}

**Step two:** find the value of 'r' corresponding to our term.

{eq}6 - r = 2 \\ r = 4 {/eq}

**Step three:** put the value of 'r' in the binomial coefficient to find its value.

{eq}= \binom{6}{4} \\ =\dfrac{6!}{4! 2!} \\ = \dfrac{6.5}{2} \\ = 15 {/eq}

So, the coefficient of {eq}x^2 y^4 {/eq} term is 15.

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from Algebra II Textbook

Chapter 21 / Lesson 16