# Section – The Binomial Theorem

Before we go any further, we need to look at a special kind of combination problem where we have "indistinguishable objects". For example, suppose I have two different kinds of marbles like three green ones and four blue ones. How many different ways are there to arrange the marbles? This is just really just a special case of the combinations that we discussed in the last chapter but it's going to be important in the rest of the section so we're going to spend a little time discussing it to make sure the reasoning behind the calculations is clear.

To answer this, think of the positions, 1 through 7, that the marbles will be assigned to. There are 7C3 ways that the seven available positions can be assigned to the three green marbles. Once we've assigned those 3 positions, there's only 1 way that the four blue marbles can be assigned so the total number of ways that three green and four blue marbles can be arranged is _{7}*C*_{3} · 1 or just _{7}*C*_{3}.

# Example 1 – Identical Objects

**How many ways can two x's and one y be arranged?**

We're arranging the two letters in a group of 3 so if we assign two of the three positions to the two *x*'s that gives us _{3}*C*_{2} combinations. There's only one remaining choice for the *y* variable so the total number of combinations is

_{3}*C*_{2} · 1 = 3

# Example 2 – Identical Objects

**How many ways can two x's and one y be arranged if you assign the positions to the y instead of the x?**

Thinking in terms of the *y* instead of the *x* means we have to assign one of the 3 positions to the one *y*. There are _{3}*C*_{1} ways to do that. Again, there's only one remaining way the *x*'s can be assigned so the total number of combinations is

_{3}*C*_{1} · 1 = 3

Examples 1 and 2 illustrate an important property. Both approaches are reasonable but there can be only one answer to the question, "How many ways can two x's and one y be arranged.", so it's good that we got the same result with both methods. If we hadn't we would have had to go back to the beginning and rethink our reasoning.

## The Binomial Theorem

Suppose that I asked you to multiply out (*x* + *y*)^{8}. You could do it using techniques that you learned in algebra by writing out

(*x* + *y*)(*x* + *y*)(*x* + *y*)(*x* + *y*)(*x* + *y*)(*x* + *y*)(*x* + *y*)(*x* + *y*)

and then multiplying out all of the factors two at a time but that's a lot of work. There's going to be a huge number of terms that you have to manipulate, combine, simplify, etc. which means there are going to be lots of opportunities to make a little mistake that ruins your final result. If you did all of that work, the final result would look like

*x*^{8} + 8*x*^{7}*y* + 28*x*^{6}*y*^{2} + 56*x*^{5}*y*^{3} + 70*x*^{4}*y*^{4} + 56*x*^{3}*y*^{5} + 28*x*^{2}*y*^{6} + 8*x*^{}*y*^{7} + *y*^{8}

When I look at the coefficients of the terms, I see a pattern. They start with 1, 8, 28, 56 and 70 and then they decrease following that exact pattern: 56, 28, 8 then 1. When mathematicians see a pattern like that, they immediately start to wonder if there's a way to predict the pattern. If we could do that, then we'd have a formula that we could use to calculate the coefficients of a "binomial expansion" like (*x* + *y*)^{8} without having to go to the effort of multiplying it out.

One way that we sometimes approach a problem like this is by looking at a smaller example like (*x* + *y*)^{3}. If I wrote that expression out, it would look like

(*x* + *y*)(*x* + *y*)(*x* + *y*)

Now suppose that I multiplied that out. The result would be

(*x* + *y*)(*x* + *y*)(*x* + *y*)

(*xx* + *xy* + *yx* + *yy*)(*x* + *y*)

*xxx* + *xyx* + *yxx* + *yyx* + *xxy* + *xyy* + *yxy* + *yyy*

That's a strange way to write it out because I didn't use exponents or combine the like terms but I did it that way for a reason. Notice how every term has 3 variables in it and every permutation of *x*'s and *y*'s appears in the expression.

Now suppose I wanted to know what the coefficient of the *x*^{2}*y* term was going to be. That's the same as asking how many of the terms in my expression have exactly 2 *x*'s or, "How many ways can two *x*'s and one *y* be arranged?" But wait! We already answered that question in Example 1: _{3}*C*_{2} or 3. Because there are 3 terms in the expression that simplify to *x*^{2}*y* that means that the coefficient of that term has to be 3.

How about the *xy*^{2} term? We can get its coefficient with the same argument. Any term with one *x* and two *y*'s will simplify to *xy*^{2}. There are _{3}*C*_{1} ways that one *x* and two *y*'s can be arranged which means there are 3 terms that simplify to *xy*^{2} so the coefficient of that term has to be 3.

If we look at the entire expansion of (*x* + *y*)^{3}, it's going to have four terms in it. The table below lists those terms and their coefficients using our combinations argument.

Term | Combination | Coefficient |
---|---|---|

x^{3} | How many ways can three x's and zero y's be arranged? | _{3}C_{3} = 1 |

x^{2}y | How many ways can two x's and one y be arranged? | _{3}C_{2} = 3 |

xy^{2} | How many ways can one x and two y's be arranged? | _{3}C_{1} = 3 |

y^{3} | How many ways can zero x's and three y's be arranged? | _{3}C_{0} = 1 |

Take a close look at the combinations in the right hand column and you should start to see a pattern. The first number in the combination is always the exponent in the original expression (3 from (*x* + *y*)^{3}) and the second number is the exponent of the *x*. In other words

coefficient = _{(exponent in the original exponent)}C_{(exponent of the x in the term)}

The Binomial Theorem summarizes our arguments into one equation.

## The Binomial Theorem

(*x* + *y*)^{n} = $ x^n + _nC_{n-1} x^{n-1}y^1 + _nC_{n-2} x^{n-2}y^2 + ... + _nC_1 x^1y^{n-1} + _nC_0 y^n$

# Example 3 – Binomial Coefficients

**What's the coefficient of the x^{3}y^{4} term in the expansion of (x + y)^{7}?**

According to the Binomial Theorem

$$\textrm{coefficient}=\begin{pmatrix}\textrm{original exponent}\\\textrm{exponent of the x}\end{pmatrix}=\begin{pmatrix}7\\3\end{pmatrix}=\frac{7!}{3! 4!} = 35$$# Example 4 – Binomial Coefficients

**What's the coefficient of the sixth term in the expansion of ( x + y)^{12}?**

This one is a little tricky because we have to figure out what the *x*'s exponent in the sixth term is going to be. The exponent on the first term is 12 so if you count up from there, the second term's exponent is 11, the third term's is 10, . . ., the sixth's term is 7 which means we're looking for the coefficient of the *x*^{7}*y*^{5} term.

# Example 5 – Binomial Expansions

**What is the expansion of ( x + y)^{5}?**

The Binomial Theorem lets us skip over multiplying out 5 *x* + *y*'s and just calculate the coefficients of each term.

Term | Coefficient |

x^{5} | _{5}C_{5} = 1 |

x^{4}y | _{5}C_{4} = 5 |

x^{3}y^{2} | _{5}C_{3} = 10 |

x^{2}y^{3} | _{5}C_{2} = 10 |

xy^{4} | _{5}C_{1} = 5 |

y^{5} | _{5}C_{0} = 1 |

That makes the expansion (*x* + *y*)^{5} = *x*^{5} + 5*x*^{4}*y* + 10*x*^{3}*y*^{2} + 10*x*^{2}*y*^{3} + 5*xy*^{4} + *y*^{5}.

Example 5 illustrates why the Binomial Theorem is such a useful tool. Multiplying out (x + y)5 would have taken me several minutes and I've been doing this for a while so I can crank that sort of thing out pretty fast. By using the combination-based formula that the Binomial Theorem gives us for the coefficients, I can jump right to the final answer in a fraction of the time.

# Example 6 – Binomial Expansions

**What is the expansion of ( x + 1)^{5}?**

Don't be thrown by the fact that I replaced the *y* with a 1. We can do this using the exact same table/pattern that we used in Example 5. We're just going to use a 1 in place of the *y*.

Term | Coefficient |

x^{5} | _{5}C_{5} = 1 |

x^{4} · 1 | _{5}C_{4} = 5 |

x^{3} · 1^{2} | _{5}C_{3} = 10 |

x^{2} · 1^{3} | _{5}C_{2} = 10 |

x · 1^{4} | _{5}C_{1} = 5 |

1^{5} | _{5}C_{0} = 1 |

That makes the expansion

$$(x + 1)^5 = x^5 + 5x^4 \cdot 1 + 10x^3 \cdot 1^2 + 10x^2 \cdot 1^3 + 5x^4 + 1^5$$

$$(x + 1)^5 = x^5 + 5x^4 + 10x^3 + 10x^2 + 5x^4 + 1$$

# Example 7 – Binomial Expansions

**What is the expansion of (3 x - 2)^{6}?**

The process here is almost identical to the previous examples but I'm going to use 3*x* instead of just *x* and -2 instead of *y*.

Term | Coefficient |

(3x)^{6} | _{6}C_{6} = 1 |

(3x)^{5} · (-2) | _{6}C_{5} = 6 |

(3x)^{4} · (-2)^{2} | _{6}C_{4} = 15 |

(3x)^{3} · (-2)^{3} | _{6}C_{3} = 20 |

(3x)^{2} · (-2)^{4} | _{6}C_{2} = 15 |

(3x)^{1} · (-2)^{5} | _{6}C_{1} = 6 |

(3x)^{0} · (-2)^{6} | _{6}C_{0} = 1 |

That makes the expansion

$$(3x - 2)^6 = (3x)^6 + 6(3x)^5 \cdot (-2) + 15(3x)^4 \cdot (-2)^2 + 20(3x)^3 \cdot (-2)^3 + 15(3x)^2 \cdot (-2)^4 + 6(3x)^1 \cdot (-2)^5 + 1(3x)^0 \cdot (-2)^6$$

$$(3x - 2)^6 = 729x^6 - 2916x^5 + 4860x^4 - 4320x^3 + 2160x^2 - 576x + 64$$