7.6: The Binomial Theorem
Learning objectives
 Work with combinations.
 Generate rows of Pascal’s triangle.
 Use the binomial theorem to expand polynomials, and to identify terms for a given polynomial.
 Use the binomial theorem to calculate the probability of success or failure in a Bernoulli trial.
Introduction
In this lesson we will develop and prove the Binomial Theorem. This theorem tells us about the relationship between two forms of a certain type of polynomial: the factored, binomial form (x + y)^{n}, and the expanded form of this polynomial. To understand the theorem, we will first revisit a topic you have seen previously, factorials, and we will discuss several other ideas needed to develop the theorem. Then we will prove the theorem using induction. Finally, we will use the theorem to expand polynomials and to solve a certain type of probability problem.
Preliminary ideas: factorials, combinations, and Pascal’s Triangle
Recall that a factorial of a positive integer n is the product of n, and all of the positive integers less than n. We write this as n! = n(n  1)(n  2) .. (3) (2) (1).
In order to develop the binomial theorem, we need to look at a related idea: combinations. If you have studied probability, you may be familiar with combinations and permutations. A combination is the number of ways you can choose r objects from a group of n objects, if the order of choosing does not matter. An example will help clarify the idea of a combination:
 In a class of 20 students, 3 students are going to be chosen to form a committee to plan a fieldtrip. How many possible committees are there?
 To answer this question, we need to figure out how many ways we can choose groups of 3 students from the 20 on the class. The order of choosing does not matter. That is, if I choose Amy, Juan, and Nina, it is the same as choosing Juan, then Amy, then Nina, or any other ordering of the three students.
In general, we can find the number of combinations of r objects chosen from n objects by the following:
\begin{align*}_{n}C_{r} = \binom{n} {r} = \frac{n!} {r!(n  r)!}\end{align*} 

(Note that there are two different symbols for combinations: \begin{align*}_{n}C_{r}\end{align*} and \begin{align*}\binom{n} {r}\end{align*} You can use either one, though \begin{align*}_{n}C_{r}\end{align*} is what is used on the TI83/84.)
 Therefore the number of combinations of 3 people from 20 people in the class is
\begin{align*}\frac{20!} {3!17!} = 1140\end{align*} 

A note about calculating combinations: for small numbers, it is not too difficult to calculate by hand. For example, \begin{align*}\binom{7} {4} = \frac{7!} {4!3!} = \frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} {4 \cdot 3 \cdot 2 \cdot 1 \cdot 3 \cdot 2 \cdot 1} = \frac{7 \cdot 6 \cdot 5} {3 \cdot 2 \cdot 1} = 7 \cdot 5 = 35\end{align*}.
Canceling factors in the numerator and denominator simplifies the calculation. However, you can always use a graphing calculator. For example, to find \begin{align*}\binom{20} {3} \end{align*} , press: 20 <TI font_MATH> and then move right to the PRB menu. Press 3. This takes you back to the main screen. You should see 20 \begin{align*}_{n}C_{r}\end{align*}. Now press 3 >,TI font_ENTER>. You should see the answer, 1140.
There is a particular pattern in combinations that is seen in the expansion of polynomials of the form (x + y)^{n}. This pattern is most commonly displayed in a triangle:
\begin{align*}\binom{0} {0}\end{align*} 

\begin{align*}\binom{1} {0} \binom{1} {1}\end{align*} 

\begin{align*}\binom{2} {0} \binom{2} {1} \binom{2} {2}\end{align*} 

\begin{align*}\binom{3} {0} \binom{3} {1} \binom{3} {2} \binom{3} {3}\end{align*} 

\begin{align*}\binom{4} {0} \binom{4} {1} \binom{4} {2} \binom{4} {3} \binom{4} {4}\end{align*} 

\begin{align*}\vdots \ \ \ \ \ \ \vdots \ \ \ \ \ \ \vdots \ \ \ \ \ \ \vdots \ \ \ \ \ \ \vdots \ \ \ \ \ \ \vdots \ \ \ \ \ \ \end{align*} 

If we evaluate the combinations in the triangle, we get:
This triangle is referred to as Pascal’s triangle, named after mathematician Blaise Pascal, although other mathematicians before him worked with these numbers. The numbers in the triangle can be used to generate more rows: notice that if you add two consecutive numbers, you get the number between and below them in the next row:
We can generalize this pattern as follows: \begin{align*}\binom{n} {r  1} + \binom{n} {r} = \binom{n + 1} {r}\end{align*}.
We can prove this statement directly:

\begin{align*}\binom{n} {r  1} + \binom{n} {r}= \frac{n!} {(r  1)!(n  (r  1))!} + \frac{n!} {r!(n  r)!}\end{align*} 






\begin{align*}= \frac{n!} {(r  1)!(n  r + 1))!} + \frac{n!} {r!(n  r)!}\end{align*} \begin{align*}= \frac{rn!} {r(r  1)!(n  r + 1)!} + \frac{n!} {r!(n  r)!}\end{align*} Multiply the first fraction by \begin{align*}\frac{r}{r}\end{align*}. Substitute \begin{align*} \,\! r(r1)!\end{align*} with \begin{align*}\,\! r!\end{align*} \begin{align*}= \frac{rn!} {r!(n  r + 1)!} + \frac{n!(n  r + 1)} {r!(n  r)!(n  r +1)}\end{align*} Multiply the second fraction by \begin{align*}\frac{(n  r + 1)}{(nr+1)}\end{align*}. Substitute \begin{align*}\,\! (n  r + 1)(n  r)!\end{align*} with \begin{align*}\,\! (n  r + 1)!\end{align*} \begin{align*}= \frac{rn!} {r!(n  r + 1)!} + \frac{n!(n  r +1)} {r!(n  r + 1)!}\end{align*} Now the fractions have a common denominator. Add the fractions and simplify the numerator: \begin{align*}= \frac{rn! + n!(n  r + 1)} {r!(n  r + 1)!}\end{align*} \begin{align*}= \frac{n!(r + n  r + 1)} {r!(n  r + 1)!} = \frac{n!(n + 1)} {r!(n  r + 1)!}\end{align*} Factor out \begin{align*}n!\end{align*}. Simplify \begin{align*}r + n  r + 1 = n + 1\end{align*} \begin{align*}= \frac{(n + 1)!} {r!(n  r + 1)!} = \frac{(n + 1)!} {r!(n + 1  r)!} = \binom{n + 1} {r}\end{align*}.






This proves that we can always generate rows of Pascal’s triangle. Further, the fact that \begin{align*}\binom{n} {r  1} + \binom{n} {r} = \binom{n + 1} {r}\end{align*} will be used below to prove the Binomial Theorem.
Binomial Expansions
To expand a binomial is to multiply all of the factors. The resulting polynomial is in standard form. For example:
(x + y)^{2} = (x + y) (x + y) = x^{2} + xy + xy + y^{2} = x^{2} + 2xy + y^{2}
If we expand (x + y)^{3} , we get:
(x + y)^{3}  = (x + y) (x + y) (x + y)  

= (x + y) (x^{2} + 2xy + y^{2})  
= x^{3} + 2x^{2}y + xy^{2} + x^{2}y + 2xy^{2} + y^{3}  
= x^{3} + 3x^{2}y + 3xy^{2} + y^{2} 
Notice that the coefficients of each polynomial correspond to a row of Pascal’s triangle. Also notice that the exponents of x descend, and the exponents of y ascend with each term. These are key aspects of the Binomial Theorem.
The Binomial Theorem
The binomial theorem can be stated using a summation:
\begin{align*}(x + y)^n = \sum_{r = 0}^n \left (\binom{n} {r} x^{n  r} y^r \right)\end{align*} 

This is a very succinct way of summarizing the pattern in a binomial expansion. Let’s return to (x + y)^{3} to see how the theorem works.
\begin{align*}(x + y)^3\end{align*}  \begin{align*}= \binom{3} {0}x^{3  0} y^0 + \binom{3} {1}x^{3 1}y^1 + \binom{3} {2}x^{3  2}y^2 + \binom{3} {3}x^{3 3}y^3\end{align*}  

\begin{align*}= 1x^3 \cdot 1 + 3x^2y + 3xy^2 + 1 \cdot x^0 \cdot y^3\end{align*}  
\begin{align*}= x^3 + 3x^2y + 3xy^2 + y^3\end{align*} 
Again, the exponents on x descend from 3 to 0. The exponents on y ascend from 0 to 3. The coefficients on the terms correspond to row 3 of Pascal’s triangle. These coefficients are, not surprisingly, referred to as binomial coefficients!
Given this theorem, we can expand any binomial without having to multiply all of the factors. We can also identify a particular term in an expansion. Before we solve these kinds of problems, we will prove the theorem:
The Binomial Theorem: \begin{align*}(x + y)^n = \sum_{r = 0}^n \left(\binom{n} {r} x^{n  r}y^r \right)\end{align*}
Proof by induction:
 1. The base case

 If n = 1, then the binomial is (x + y)^{1} = x + y = 1x + 1y. The coefficients correspond to row 1 (physically the second row) of the triangle.


 Note: we have already observed that the theorem holds for n = 2 and n = 3.

 2. The inductive hypothesis:

 Assume that \begin{align*}(x + y)^k = \sum_{r = 0}^k \left(\binom{k} {r}x^{k  r}y^r \right)\end{align*}.
 3. The inductive step: show that \begin{align*}(x + y)^{k + 1} = \sum_{r = 0}^{k + 1} \left(\binom{k + 1} {r} x^{k + 1r}y^r \right)\end{align*}


\begin{align*}(x + y)^{k + 1}= (x + y) (x + y)^k\end{align*} 


\begin{align*}= (x + y) (x^k + \sum_{r = 1}^{k  1} \binom{k} {r}x^{k  r} y^r + y^k)\end{align*} Use the inductive hypothesis, but write out the first and last term: \begin{align*}x^k, y^k\end{align*} \begin{align*}= x^{k + 1} + y^{k + 1} + yx^k + xy^k + x \left(\sum_{r = 1}^{k  1} \binom{k} {r}x^{k  r}y^r \right) + y \left(\sum_{r = 1}^{k  1} \binom{k} {r}x^{k  r}y^r \right)\end{align*} Multiply \begin{align*}= x^{k + 1} + y^{k + 1} + yx^k + xy^k + \sum_{r = 1}^{k  1} \binom{k} {r}x^{k +1  r}y^r + \sum_{r = 1}^{k  1} \binom{k} {r}x^{k  r}y^{r + 1}\end{align*} \begin{align*}yx^k\end{align*} is \begin{align*}\binom{k} {0} x^{k0}y^{0 + 1}\end{align*} and \begin{align*}= x^{k + 1} + y^{k + 1} + \sum_{r = 1}^{k  1} \binom{k} {r}x^{k +1  r}y^r + \sum_{r = 0}^{k  1} \binom{k} {r}x^{k  r}y^{r + 1}\end{align*} \begin{align*}xy^k\end{align*} is \begin{align*}\binom{k} {1} x^{k + 11}y^1\end{align*} \begin{align*}= x^{k + 1} + y^{k + 1} + \sum_{r = 1}^{k  1} \binom{k} {r}x^{k +1  r}y^r + \sum_{j = 1}^{k } \binom{k} {j1}x^{k + 1 j}y^j \end{align*} Let j = r+1 and rewrite the sum. \begin{align*}= x^{k + 1} + y^{k + 1} + \sum_{r = 1}^{k  1} \binom{k} {r}x^{k +1  r}y^r + \sum_{r = 1}^{k} \binom{k} {r1}x^{k + 1  r}y^r\end{align*} Then rewrite the sum again with r. \begin{align*}= x^{k + 1} + y^{k + 1} + \sum_{r = 1}^{k} \left [\binom{k} {r} + \binom{k} {r1} \right ] x^{k +1  r}y^{r}\end{align*} It was proven above:\begin{align*}\binom{n} {r1}+{n} {r1}=\binom{n+1} {r}\end{align*} \begin{align*}= x^{k + 1} + y^{k + 1} + \sum_{r = 1}^{k } \binom{k+1} {r}x^{k +1  r}y^r + \sum_{r = 0}^{k+1} \binom{k+1} {r1}x^{k + 1  r}y^r\end{align*}




 This proves the Binomial Theorem. Now we can use the theorem to expand binomials, and to identify terms and coefficients in an expansion.
Expansion, and Terms and Coefficients in a Binomial Expansion
As noted above, the binomial theorem allows us to expand binomials. So far we have only seen the binomial (x + y)^{n} , but the two terms in the binomial could be any monomials.
Example 1: Use the binomial theorem to expand each polynomial:

a.(2x + a)^{4} b. (x  3)^{5}
Solution:
 a.(2x + a)^{4}
= 1(2x)^{4} (a)^{0} + 4(2x)^{3} (a)^{1} + 6(2x)^{2}(a)^{2} + 1(2x)^{0} (a)^{4}  

= 16x^{4} + 32x^{3}a + 24x^{2}a^{2} + 8x'a^{3} + a^{4} 

 Note that it is easier to simply use the numbers from the appropriate row of the triangle than to write out all of the coefficients as combinations. However, if n is large, it may be easier to use the combinations.
 b. (x  3)^{5}
= 1(x)^{5} (3)^{0} + 5(x)^{4} (3)^{1} + 10(x)^{3}(3)^{2} + 10(x)^{2}(3)^{3} + 5(x)^{1} (3)^{4} + 1(x)^{0} (3)^{5}  

= x^{5}  1x^{4}a + 90x^{3}  270x^{2} + 405x  243 

 Notice that in this expansion, the terms alternate signs. This is the case because the second term in the binomial is 3. When expanding this kind of polynomial, be careful with your negatives!
We can also use the Binomial Theorem to identify a particular term or coefficient.
Example 2: Identify the 3^{rd} term of the expansion of (2x + 3)^{6}.
Solution: The 3^{rd} term is \begin{align*}\mathit \ \binom{6}{2} (2x)^{4}3^{2}=15\cdot 16x^{4} \cdot 9 = 2160x^{4}\end{align*}
 Keep in mind that row 6 of Pascal’s triangle starts with \begin{align*}\mathit \ \binom{6}{0}\end{align*} , so the coefficient of the third term in the expansion is \begin{align*}\mathit \ \binom{6}{2}\end{align*}. Also keep in mind that the first term includes (2x)^{6} and 3^{0} , so the third term includes (2x)^{4} and 3^{2}.
Finding a term in an expansion can be used to answer a particular kind of probability question. Consider an experiment, in which there are two possible outcomes, such as flipping a coin. If we flip a coin over and over again, this is referred to as a Bernoulli trial. In each flip (“experiment”), the probability of getting a head is 0.5, and the probability of getting a tail is 0.5. (Note: this is true for flipping a coin, but not for other situations. That is, it’s not always “5050 chance!) Now say we flip a coin 25 times. What is the probability of getting exactly 10 heads?
The answer to this question is a term of a binomial expansion. That is, the probability of getting 10 heads from 25 coin tosses is: \begin{align*} \binom{25}{10} (0.5)^{10}(0.5)^{15}\approx 0.0974\end{align*}, or about a 10% chance. Now let’s consider an example that is not a “5050” situation.
Example 3: At the carnival, you decide to play a game of chance. You buy 15 tickets for the game. You have a 75% chance of winning each time you play the game. What is the probability that you will win exactly 8 of the 15 games?
Solution: About 4%
 For each of the 15 games, there is 75% chance of winning, and a 25% chance of losing. So the probability of exactly 8 wins is \begin{align*}\mathit \ \binom{15}{8}(.75)^{8}(.25)^{7}\approx 0.039\end{align*}, or about a 4% chance. It is important to note that this is only the probability of winning exactly 8 games. If for example, you need to win 8 or more games in order to win a certain prize, the probability of winning would be higher, as it would be the sum of the probabilities of winning 8, 9, 10, 11, 12, 13, 14, and 15 games.
Summary
In this lesson we have developed, proved, and used the Binomial Theorem. This theorem is useful because it allows us to expand polynomials without having to do many steps of multiplication of factors. It is also useful because it allows us to identify a term of an expansion, which can be used to solve certain probability problems. In your study of calculus, you will also find that binomial expansions are useful for approximating certain polynomials.
Points to Consider
 Pascal’s triangle is named after mathematician Blaise Pascal, even though he was not the first person to work with these numerical relationships. Why might this be the case?
 What happens to a binomial expansion if n is not a positive integer?
 Why is the probability of winning k out of n games not k/n?
Review Questions
 Write out the combination and find its value: a. \begin{align*} _{8}C_{5}\end{align*} b. \begin{align*} _{6}C_{3}\end{align*}
 In a class of 200 students, 25 will be chosen randomly to participate in a research study. How many possible groups of 25 students can be chosen? (Hint: use a calculator!)
 Expand: \begin{align*}(x+3a)^{4}\end{align*} =
 Expand: \begin{align*}(y + \frac{1}{2})^5=\end{align*}
 Expand: \begin{align*}(2xa)^{6}\end{align*}=
 Find the 3^{rd} term in the expansion \begin{align*}(3x+2a)^{9}\end{align*}.
 Find the 7^{th} term in the expansion of \begin{align*}(4x  \frac{1}{2}a)^{10}\end{align*}.
 A die is rolled 10 times. What is the probability of rolling exactly four 4’s? (Hint: the probability of rolling a 4 is 1/6. The probability of not rolling a 4 is 5/6.)
 The local TV station forecasts a 30% chance of rain every day for the next week. What is the probability that it will rain on exactly 6 out of the next 7 days?
 Consider the following situation: a basketball player is going to attempt to make 20 free throws. She is assuming that she has an 80% chance of making each shot. What is the probability that she will make exactly 19 out of 20 shots? Two students were discussing this problem. Student A says that the problem is a Bernoulli trial, and that the probability is \begin{align*}\binom{20}{19}(.8)^{19}(.2)^1\end{align*} . Student B disagrees, and says that the situation is not a Bernoulli trial. What reasoning might student B use to support his argument?
Review Answers
 a.\begin{align*}\frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot 3 \cdot 2 \cdot 1} = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = 8 \cdot 7 = 56\end{align*} b.\begin{align*}\frac{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 }{3 \cdot 2 \cdot 1 \cdot 3 \cdot 2 \cdot 1} = \frac{6 \cdot 5 \cdot 5}{3 \cdot 2 \cdot 1} = 5 \cdot 4 = 20\end{align*}
 \begin{align*}_{200}C_{25} \approx 4.52 \times 10^{11} \end{align*}
 x^{4} + 12x^{3}a + 54x^{2}a^{2} + 108xa^{3} + 81a^{4}
 \begin{align*}y^5 + \frac{5}{2}y^4 + \frac{5}{2}y^3 + \frac{5}{4}y^2 + \frac{5}{16}y + \frac{1}{32} \end{align*}
 \begin{align*}64x^6 + 19.2x^5 + 2.4x^4 + 0.16x^3 + 0.006x^2 + 0.00012x^5 + 0.000001x^6 \end{align*}
 \begin{align*}\binom{9}{3}(3x)^7 (2a)^2 = 8.748x^7a^4\end{align*}
 \begin{align*}\binom{10}{7}(4x)^4(\frac{1}{6}a)^6 = \frac{160}{243}x^4a^6\end{align*}
 About 5.4%
 About 0.36%
 If the situation is a Bernoulli trial, then the probabilities of success and failure are the same with each shot. The player has assumed this, but this might not be the case. For example, if she improves as she shoots, the probability of success increases. Or, if she gets tired, distracted, or demotivated, the probability of success might decrease.
Vocabulary
 Bernoulli trial
 A Bernoulli trial is a series of experiments in which each experiment has two possible outcomes, usually referred to as “success” and “failure.”
 Binomial
 A binomial is a polynomial with two terms.
 Combination
 A combination refers to choosing one or more objects from a set when the order of choosing does not matter.
 Factorial
 Factorial refers to the product of the positive integers from 1 to some value n:1 × 2 × 3 × 4....×(n1)×n=n!
 Monomial
 A monomial is a term of the form ax^{n} , where a is a real number and n is a whole number.
 Polynomial
 A polynomial is a sum of monomials.
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