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# Common Core Standards: Math

# Math.CCSS.Math.Content.HSA-APR.C.5

**5. Know and apply the Binomial Theorem for the expansion of ( x + y)^{n} in powers of x and y for a positive integer n, where x and y are any numbers, with coefficients determined for example by Pascal's Triangle.**

The Binomial Theorem and Pascal's Triangle are very important concepts for two reasons. First, they're capitalized, and when words are capitalized, you know they're a big deal. Second, both the Binomial Theorem and Pascal's Triangle make expanding polynomials way easier.

Students should be able to expand (*x + y*)^{n} using Pascal's Triangle, where *x* and *y* can be anything. Pascal's Triangle provides the coefficients for the expansion, and the Binomial Theorem explains Pascal's Triangle. They go together like ramma lamma lamma ke ding a de dinga dong…whatever that means.

The Binomial Theorem can be explained using combinations, usually because they're the easiest to visualize and explain. For example, flipping a coin is a classic example since there are two sides.

How many combinations are possible with 1 heads and 2 tails? We could have HTT, THT, and TTH. There are three possible combinations. We can assign values to our total number of trials (*n* = 3) and our total number of desired outcome (heads, for example, would be *k* = 1).

The notation _{n}C_{k} is a way of saying, "The number of different combinations we can have if, out of *n* trials, our desired outcome occurs *k *times." For example, we can use _{n}C_{k} if we flip a coin *n* times and we want to know how many different ways we can have *k* heads (or *k* tails, whichever we want). We can calculate _{n}C_{k} using the formula

Those exclamation points aren't just because we're excited about *n* and *k*. They're mathspeak for "factorial," which translates to *n* × (*n* – 1) × (*n* – 2) × … × 2 × 1. But you can be excited about *n* and *k* too.

In our example of three coin flips (*n* = 3) and 1 heads (*k* = 1), this equals

There are 3 possible ways to get 1 heads out of 3 coin flips.

We can use this same logic when expanding (*x* + *y*)^{n}. How many ways can we get 3 *x*'s and 2 *y*'s, for example? (Note that there are the same number of ways to get 2 *x*'s and 3 *y*'s.) This can be written as _{5}C_{3}, which equals

Let's now think about all ways of getting *k* *x*'s and (*n* – *k*) *y*'s out of *n* outcomes.

When *n* = 0, we can have _{0}C_{0}, which equals 1. Our values for *n* = 2 are _{1}C_{0} and _{1}C_{1}, both of which equal 1 as well. For *n* = 2, _{2}C_{0} and _{2}C_{2} equal 1, but _{2}C_{1} equals 2. Students should know that continuing with these calculations would get us Pascal's Triangle, in which each number is the sum of the two numbers above it.

Let's use Pascal's Triangle to expand (*x* + *y*)^{5}.

(*x* + *y*)^{5} = *x*^{5} + 5*x*^{4}*y* + 10*x*^{3}*y*^{2} + 10*x*^{2}*y*^{3} + 5*xy*^{4} + *y*^{5}

The coefficients of each term correspond to the row of Pascal's Triangle in which *n *= 5. The exponents of the *x*'s start from *n* and decrease to 0, while the orders of the *y*'s start at 0 and increase until we get to *n*. Pretty awesome, right?