To find the coefficient of the term

x^{2} y^{4}

of the binomial expansion, we should recall the Binomial Theorem. It states that for every positive integer

n,

we can expand the expression

(a + b)^{n}

by using the numbers in the

n^{th}

row of Pascal's Triangle.

(a + b)^{n} = P_{0} a^{n} b^{0} + P_{1} a^{n - 1} b^{1} + ⋮ + P_{n - 1} a^{1} b^{n - 1} + P_{n} a^{0} b^{n}

In the above formula,

P_{0},

P_{1},

\dots,

P_{n}

are the numbers in the

n^{th}

row of Pascal's Triangle.

$(a + b)^{n}$ $=$ $P_{0} a^{n} b^{0}$ $+$ $P_{1} a^{n - 1} b$ $+$ $\dots$ $+$ $P_{n - 1} a b^{n - 1}$ $+$ $P_{n} a^{0} b^{n}$

It can be shown that the $i^{th}$ coefficient of the $n^{th}$ row of the Pascal's Triangle equals $_{n} C_{i} .$ Therefore, we can restate the binomial theorem using combinations.

$(a + b)^{n}$ $=$ $_{n} C_{0} a^{n} b^{0}$ $+$ $_{n} C_{1} a^{n - 1} b$ $+$ $\dots$ $+$ $_{n} C_{n - 1} a b^{n - 1}$ $+$ $_{n} C_{n} a^{0} b^{n}$

Next, let's recall the formula to calculate the combination

_{n} C_{x} .

_{n} C_{x} \Leftrightarrow \frac{n !}{x ! ( n - x ) !}

We will use this to substitute the values of the combinations in the formula for the binomial expansion.

$(a + b)^{n} = P_{0} a^{n} b^{0} + P_{1} a^{n - 1} b^{1} + \dots + P_{n - 1} a^{1} b^{n - 1} + P_{n} a^{0} b^{n}$
$(x + 2 y)^{6} = 1 x^{6} (2 y)^{0} + 6 x^{5} (2 y)^{1} + 15 x^{4} (2 y)^{2} + 20 x^{3} (2 y)^{3} + 15 x^{2} (2 y)^{4} + 6 x^{1} (2 y)^{5} + 1 x^{0} (2 y)^{6}$

Exercise