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# 9.4: Binomial Theorem

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Skills to Develop

• Evaluate expressions involving factorials.
• Calculate binomial coefficients.
• Expand powers of binomials using the binomial theorem.

## Factorials and the Binomial Coefficient

We begin by defining the factorial25 of a natural number $$n$$, denoted $$n!$$, as the product of all natural numbers less than or equal to $$n$$.

$$n !=n(n-1)(n-2) \cdots 3 \cdot 2 \cdot 1$$

For example,

$$\begin{array}{l}{7 !=7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=5,040 \quad\color{Cerulean}{Seven\:factorial}} \4pt] {5 !=5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=120\quad\quad\quad\quad\:\color{Cerulean}{Five\:factorial}} \\[4pt] {3 !=3 \cdot 2 \cdot 1=6\quad\quad\quad\quad\quad\quad\:\:\:\quad\color{Cerulean}{Three\:factorial}} \\[4pt] {1 !=1=1\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\color{Cerulean}{One\:factorial}}\end{array}$$ We define zero factorial26 to be equal to $$1$$, $$0 !=1 \quad\color{Cerulean}{Zero\:factorial}$$ The factorial of a negative number is not defined. On most modern calculators you will find a factorial function. Some calculators do not provide a button dedicated to it. However, it usually can be found in the menu system if one is provided. The factorial can also be expressed using the following recurrence relation, $$n !=n(n-1) !$$ For example, the factorial of $$8$$ can be expressed as the product of $$8$$ and $$7!$$: \begin{aligned} 8 ! &=8 \cdot \color{Cerulean}{7 !} \\[4pt] &=8 \cdot \color{Cerulean}{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} \\[4pt] &=40,320 \end{aligned} When working with ratios involving factorials, it is often the case that many of the factors cancel. Example $$\PageIndex{1}$$ Evaluate: $$\frac{12 !}{6 !}$$. Solution \begin{aligned} \frac{12 !}{6 !}&=\frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot \color{Cerulean}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}}{\color{Cerulean}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}}\\[4pt] &=\frac{12\cdot11\cdot10\cdot9\cdot8\cdot7 \cdot \cancel{6!}}{\cancel{6!}} \\[4pt] &=12\cdot11\cdot10\cdot9\cdot8\cdot7\\[4pt] &=665,280\end{aligned} Answer $$665,280$$ The binomial coefficient27, denoted $$_{n} C_{k}=\left( \begin{array}{l}{n} \\[4pt] {k}\end{array}\right)$$, is read “$$n$$ choose $$k$$” and is given by the following formula: $$_{n} C_{k}=\left( \begin{array}{l}{n} \\[4pt] {k}\end{array}\right)=\frac{n !}{k !(n-k) !}$$ This formula is very important in a branch of mathematics called combinatorics. It gives the number of ways $$k$$ elements can be chosen from a set of $$n$$ elements where order does not matter. In this section, we are concerned with the ability to calculate this quantity. Example $$\PageIndex{2}$$ Calculate $$\left( \begin{array}{l}{7} \\[4pt] {3}\end{array}\right)$$. Solution: Use the formula for the binomial coefficient, $$\left( \begin{array}{l}{n} \\[4pt] {k}\end{array}\right)=\frac{n !}{k !(n-k) !}$$ where $$n = 7$$ and $$k = 3$$. After substituting, look for factors to cancel. \begin{aligned} \left( \begin{array}{l}{7} \\[4pt] {3}\end{array}\right) &=\frac{7 !}{3 !(7-3) !} \\[4pt] &=\frac{7 !}{3 ! 4 !} \\[4pt] &=\frac{7\cdot6\cdot5\cdot\cancel{\color{Cerulean}{4!}}}{3! \cancel{\color{Cerulean}{4!}}} \\[4pt]&=\frac{210}{6} \\[4pt] &=35\end{aligned} Answer: $$35$$ Check the menu system of your calculator for a function that calculates this quantity. Look for the notation $$_{n} C_{k}$$ in the probability subsection. Exercise $$\PageIndex{1}$$ Calculate $$\left( \begin{array}{l}{8} \\[4pt] {5}\end{array}\right)$$. Answer $$56$$ http://www.youtube.com/v/Rpb8KD1HQGc Consider the following binomial raised to the $$3^{rd}$$ power in its expanded form: $$(x+y)^{3}=x^{3}+3 x^{2} y+3 x y^{2}+y^{3}$$ Compare it to the following calculations, $$\left( \begin{array}{l}{3} \\[4pt] {0}\end{array}\right)=\drac{3 !}{0 !(3-0) !}=\frac{3 !}{1 \cdot 3 !}=1$$ $$\left( \begin{array}{l}{3} \\[4pt] {1}\end{array}\right)=d\frac{3 !}{1 !(3-1) !}=\frac{3 \cdot 2 !}{1 \cdot 2 !}=3$$ $$\left( \begin{array}{l}{3} \\[4pt] {2}\end{array}\right)=\dfrac{3 !}{2 !(3-2) !}=\frac{3 \cdot 2 !}{2 !}=3$$ $$\left( \begin{array}{c}{3} \\[4pt] {3}\end{array}\right)=\dfrac{3 !}{3 !(3-3) !}=\frac{3 !}{3 ! 0 !}=1$$ Notice that there appears to be a connection between these calculations and the coefficients of the expanded binomial. This observation is generalized in the next section. ## Binomial Theorem Consider expanding $$(x+2)^{5}$$: $$(x+2)^{5}=(x+2)(x+2)(x+2)(x+2)(x+2)$$ One quickly realizes that this is a very tedious calculation involving multiple applications of the distributive property. The binomial theorem28 provides a method of expanding binomials raised to powers without directly multiplying each factor: $$(x+y)^{n}=\left( \begin{array}{c}{n} \\[4pt] {0}\end{array}\right) x^{n} y^{0}+\left( \begin{array}{c}{n} \\[4pt] {1}\end{array}\right) x^{n-1} y^{1}+\left( \begin{array}{c}{n} \\[4pt] {2}\end{array}\right) x^{n-2} y^{2}+\ldots+\left( \begin{array}{c}{n} \\[4pt] {n-1}\end{array}\right) x^{1} y^{n-1}$$ More compactly we can write, $$(x+y)^{n}=\sum_{k=0}^{n} \left( \begin{array}{l}{n} \\[4pt] {k}\end{array}\right) x^{n-k} y^{k} \quad \color{Cerulean}{Binomial\:theorem}$$ Example $$\PageIndex{3}$$ Expand using the binomial theorem: $$(x + 2)^{5}$$. Solution Use the binomial theorem where $$n = 5$$ and $$y = 2$$. $$(x+2)^{5}=\left( \begin{array}{l}{5} \\[4pt] {0}\end{array}\right) x^{5} 2^{0}+\left( \begin{array}{c}{5} \\[4pt] {1}\end{array}\right) x^{4} 2^{1}+\left( \begin{array}{l}{5} \\[4pt] {2}\end{array}\right) x^{3} 2^{2}+\left( \begin{array}{l}{5} \\[4pt] {3}\end{array}\right) x^{2} 2^{3}+\left( \begin{array}{l}{5} \\[4pt] {4}\end{array}\right) x^{1}2^{4}$$ Sometimes it is helpful to identify the pattern that results from applying the binomial theorem. Notice that powers of the variable $$x$$ start at $$5$$ and decrease to zero. The powers of the constant term start at $$0$$ and increase to $$5$$. The binomial coefficients can be calculated off to the side and are left to the reader as an exercise. \begin{aligned}(x+2)^{5} &=\left( \begin{array}{c}{5} \\[4pt] {0}\end{array}\right) x^{5} 2^{0}+\left( \begin{array}{c}{5} \\[4pt] {1}\end{array}\right) x^{4} 2^{1}+\left( \begin{array}{c}{5} \\[4pt] {2}\end{array}\right) x^{3} 2^{2}+\left( \begin{array}{c}{5} \\[4pt] {3}\end{array}\right) x^{2} 2^{3}+\left( \begin{array}{c}{5} \\[4pt] {4}\end{array}\right) x^{1} 2^{4} \\[4pt] &=1 x^{5} \times 1+5 x^{4} \times 2+10 x^{3} \times 4+10 x^{2} \times 8+5 x^{1} \times 16+1 \times 1 \\[4pt] &=x^{5}+10 x^{4}+40 x^{3}+80 x^{2}+80 x+32 \end{aligned} Answer $$x^{5}+10 x^{4}+40 x^{3}+80 x^{2}+80 x+32$$ The binomial may have negative terms, in which case we will obtain an alternating series. Example $$\PageIndex{4}$$ Expand using the binomial theorem: $$(u − 2v)^{4}$$. Solution Use the binomial theorem where $$n = 4, x = u$$, and $$y = −2v$$ and then simplify each term. \begin{aligned}(u-2 v)^{4} &=\left( \begin{array}{c}{4} \\[4pt] {0}\end{array}\right) u^{4}(-2 v)^{0}+\left( \begin{array}{c}{4} \\[4pt] {1}\end{array}\right) u^{3}(-2 v)^{1}+\left( \begin{array}{c}{4} \\[4pt] {2}\end{array}\right) u^{2}(-2 v)^{2}+\left( \begin{array}{c}{4} \\[4pt] {3}\end{array}\right) u^{1}(-2v)^{3} + \left(\begin{array}{c}4 \\[4pt]4 \end{array} \right)u^{0}(-2v)^{4} \\[4pt] &=1 \times u^{4} \times 1+4 u^{3}(-2 v)+6 u^{2}\left(4 v^{2}\right)+4 u\left(-8 v^{3}\right) + 16v^{4} \\[4pt] &=u^{4}-8 u^{3} v+24 u^{2} v^{2}-32 u v^{3}+16 v^{4} \end{aligned} Answer $$u^{4}-8 u^{3} v+24 u^{2} v^{2}-32 u v^{3}+16 v^{4}$$ Exercise $$\PageIndex{2}$$ Expand using the binomial theorem: $$\left(a^{2}-3\right)^{4}$$ Answer $$a^{8}-12 a^{6}+54 a^{4}-108 a^{2}+81$$ http://www.youtube.com/v/wICbqmoa4T4 Next we study the coefficients of the expansions of $$(x + y)^{n}$$ starting with $$n = 0$$: $$(x+y)^{0}=1$$ $$(x+y)^{1}=x+y$$ $$(x+y)^{2}=x+y$$ $$(x+y)^{3}=x^{2}+2 x y+y^{2}$$ $$(x+y)^{3}=x^{3}+3 x^{2} y+3 x y^{2}+y^{3}$$ $$(x+y)^{4}=x^{4}+4 x^{3} y+6 x^{2} y^{2}+4 x y^{3}+y^{4}$$ Write the coefficients in a triangular array and note that each number below is the sum of the two numbers above it, always leaving a $$1$$ on either end. Figure 9.4.1 This is Pascal’s triangle29; it provides a quick method for calculating the binomial coefficients. Use this in conjunction with the binomial theorem to streamline the process of expanding binomials raised to powers. For example, to expand $$(x − 1)^{6}$$ we would need two more rows of Pascal’s triangle, Figure 9.4.2 The binomial coefficients that we need are in blue. Use these numbers and the binomial theorem to quickly expand $$(x − 1)^{6}$$ as follows: \begin{aligned}(x-1)^{6} &=1 x^{6}(-1)^{0}+6 x^{5}(-1)^{1}+15 x^{4}(-1)^{2}+20 x^{3}(-1)^{3}+15 x^{2}(-1)^{4}+6 x(-1)^{5} \\[4pt] &=x^{6}-6 x^{5}+15 x^{4}-20 x^{3}+15 x^{2}-6 x+1 \end{aligned} Example $$\PageIndex{5}$$ Expand using the binomial theorem and Pascal’s triangle: $$(2x − 5)^{4}$$. Solution: From Pascal’s triangle we can see that when $$n = 4$$ the binomial coefficients are $$1, 4, 6, 4$$, and $$1$$.Use these numbers and the binomial theorem as follows: \begin{aligned}(2 x-5)^{4} &=1(2 x)^{4}(-5)^{0}+4(2 x)^{3}(-5)^{1}+6(2 x)^{2}(-5)^{2}+4(2 x)^{1}(-5)^{3}+(2 x)^{0}(-5)^{4} \\[4pt] &=16 x^{4} \cdot 1+4 \cdot 8 x^{3}(-5)+6 \cdot 4 x^{2} \cdot 25+4 \cdot 2 x(-125)+1 \cdot 625 \\[4pt] &=16 x^{4}-160 x^{3}+600 x^{2}-1,000 x+625 \end{aligned} Answer: $$16 x^{4}-160 x^{3}+600 x^{2}-1,000 x+625$$ ## Key Takeaways • To calculate the factorial of a natural number, multiply that number by all natural numbers less than it: $$5! = 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 120$$. Remember that we have defined $$0! = 1$$. • The binomial coefficients are the integers calculated using the formula: \[\left( \begin{array}{l}{n} \\[4pt] {k}\end{array}\right)=\frac{n !}{k !(n-k) !}. \nonumber
• The binomial theorem provides a method for expanding binomials raised to powers without directly multiplying each factor: $(x+y)^{n}=\sum_{k=0}^{n} \left( \begin{array}{l}{n} \\[4pt] {k}\end{array}\right) x^{n-k} y^{k}\nonumber$
• Use Pascal’s triangle to quickly determine the binomial coefficients.

Exercise $$\PageIndex{3}$$

Evaluate.

1. $$6!$$
2. $$4!$$
3. $$10!$$
4. $$9!$$
5. $$\frac{6 !}{3 !}$$
6. $$\frac{8 !}{4 !}$$
7. $$\frac{13 !}{9 !}$$
8. $$\frac{15 !}{10 !}$$
9. $$\frac{12 !}{3 ! 7 !}$$
10. $$\frac{10 !}{2 ! 5 !}$$
11. $$\frac{n !}{(n-2) !}$$
12. $$\frac{(n+1) !}{(n-1) !}$$
13. (a) $$4 !+3 !$$ (b) $$(4+3) !$$
14. (a) $$4 !-3 !$$ (b) $$(4-3) !$$

1. $$720$$

3. $$3,628,800$$

5. $$120$$

7. $$17,160$$

9. $$15,840$$

11. $$n^{2} − n$$

13. a. $$30$$ b. $$5,040$$

Exercise $$\PageIndex{4}$$

Rewrite using factorial notation.

1. $$1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7$$
2. $$1 \times 2 \times 3 \times 4 \times 5$$
3. $$15 \times 14 \times 13$$
4. $$10 \times 9 \times 8 \times 7$$
5. $$13$$
6. $$8 \times 7$$
7. $$n(n-1)(n-2)$$
8. $$1 \times 2 \times 3 \times \cdots \times n \times(n+1)$$

1. $$7!$$

3. $$\frac{15 !}{12 !}$$

5. $$\frac{13 !}{12 !}$$

7. $$\frac{n !}{(n-3) !}$$

Exercise $$\PageIndex{5}$$

Calculate the indicated binomial coefficient.

1. $$\left( \begin{array}{l}{6} \\[4pt] {4}\end{array}\right)$$
2. $$\left( \begin{array}{l}{8} \\[4pt] {4}\end{array}\right)$$
3. $$\left( \begin{array}{l}{7} \\[4pt] {2}\end{array}\right)$$
4. $$\left( \begin{array}{l}{9} \\[4pt] {5}\end{array}\right)$$
5. $$\left( \begin{array}{l}{9} \\[4pt] {0}\end{array}\right)$$
6. $$\left( \begin{array}{l}{13} \\[4pt] {12}\end{array}\right)$$
7. $$\left( \begin{array}{l}{n} \\[4pt] {0}\end{array}\right)$$
8. $$\left( \begin{array}{l}{n} \\[4pt] {n}\end{array}\right)$$
9. $$\left( \begin{array}{l}{n} \\[4pt] {1}\end{array}\right)$$
10. $$\left( \begin{array}{c}{n} \\[4pt] {n-1}\end{array}\right)$$
11. $$_{10} C_{8}$$
12. $$_{5} C_{1}$$
13. $$_{12} C_{12}$$
14. $$_{10} C_{5}$$
15. $$_{n} C_{n-2}$$
16. $$_{n} C_{n-3}$$

1. $$15$$

3. $$21$$

5. $$1$$

7. $$1$$

9. $$n$$

11. $$45$$

13. $$1$$

15. $$\frac{n^{2}-n}{2}$$

Exercise $$\PageIndex{6}$$

Expand using the binomial theorem.

1. $$(4 x-3)^{3}$$
2. $$(2 x-5)^{3}$$
3. $$\left(\frac{x}{2}+y\right)^{3}$$
4. $$\left(x+\frac{1}{y}\right)^{3}$$
5. $$(x+3)^{4}$$
6. $$(x+5)^{4}$$
7. $$(x-4)^{4}$$
8. $$(x-2)^{4}$$
9. $$\left(x+\frac{2}{y}\right)^{4}$$
10. $$\left(\frac{x}{3}-y\right)^{4}$$
11. $$(x+1)^{5}$$
12. $$(x-3)^{5}$$
13. $$(x-2)^{6}$$
14. $$(x+1)^{6}$$
15. $$(x-1)^{7}$$
16. $$(x+1)^{7}$$
17. $$(5 x-1)^{4}$$
18. $$(3 x-2)^{4}$$
19. $$(4 u+v)^{4}$$
20. $$(3 u-v)^{4}$$
21. $$(u-5 v)^{5}$$
22. $$(2 u+3 v)^{5}$$
23. $$\left(a-b^{2}\right)^{5}$$
24. $$\left(a^{2}+b^{2}\right)^{4}$$
25. $$\left(a^{2}+b^{4}\right)^{6}$$
26. $$\left(a^{5}+b^{2}\right)^{5}$$
27. $$(x+\sqrt{2})^{3}$$
28. $$(x-\sqrt{2})^{4}$$
29. $$(\sqrt{x}-\sqrt{y})^{4}, x, y \geq 0$$
30. $$(\sqrt{x}+2 \sqrt{y})^{5}, x, y \geq 0$$
31. $$(x+y)^{7}$$
32. $$(x+y)^{8}$$
33. $$(x+y)^{9}$$
34. $$(x-y)^{7}$$
35. $$(x-y)^{8}$$
36. $$(x-y)^{9}$$

1. $$64 x^{3}-144 x^{2}+108 x-27$$

3. $$\frac{x^{3}}{8}+\frac{3 x^{2} y}{4}+\frac{3 x y^{2}}{2}+y^{3}$$

5. $$x^{4}+12 x^{3}+54 x^{2}+108 x+81$$

7. $$x^{4}-16 x^{3}+96 x^{2}-256 x+256$$

9. $$x^{4}+\frac{8 x^{3}}{y}+\frac{24 x^{2}}{y^{2}}+\frac{32 x}{y^{3}}+\frac{16}{y^{4}}$$

11. $$x^{5}+5 x^{4}+10 x^{3}+10 x^{2}+5 x+1$$

13. $$x^{6}-12 x^{5}+60 x^{4}-160 x^{3}+240 x^{2}-192 x+64$$

15. $$x^{7}-7 x^{6}+21 x^{5}-35 x^{4}+35 x^{3}-21 x^{2}+7 x-1$$

17. $$625 x^{4}-500 x^{3}+150 x^{2}-20 x+1$$

19. $$256 u^{4}+256 u^{3} v+96 u^{2} v^{2}+16 u v^{3}+v^{4}$$

21. $$\begin{array}{l}{u^{5}-25 u^{4} v+250 u^{3} v^{2}-1,250 u^{2} v^{3}} {+3,125 u v^{4}-3,125 v^{5}}\end{array}$$

23. $$a^{5}-5 a^{4} b^{2}+10 a^{3} b^{4}-10 a^{2} b^{6}+5 a b^{8}-b^{10}$$

25. $$\begin{array}{l}{a^{12}+6 a^{10} b^{4}+15 a^{8} b^{8}+20 a^{6} b^{12}} {+15 a^{4} b^{16}+6 a^{2} b^{20}+b^{24}}\end{array}$$

27. $$x^{3}+3 \sqrt{2} x^{2}+6 x+2 \sqrt{2}$$

29. $$x^{2}-4 x \sqrt{x y}+6 x y-4 y \sqrt{x y}+y^{2}$$

31. $$\begin{array}{l}{x^{7}+7 x^{6} y+21 x^{5} y^{2}+35 x^{4} y^{3}} {+35 x^{3} y^{4}+21 x^{2} y^{5}+7 x y^{6}+y^{7}}\end{array}$$

33. $$\begin{array}{l}{x^{9}+9 x^{8} y+36 x^{7} y^{2}+84 x^{6} y^{3}+126 x^{5} y^{4}} {+126 x^{4} y^{5}+84 x^{3} y^{6}+36 x^{2} y^{7}+9 x y^{8}+y^{9}}\end{array}$$

35. $$\begin{array}{l}{x^{8}-8 x^{7} y+28 x^{6} y^{2}-56 x^{5} y^{3}+70 x^{4} y^{4}} {-56 x^{3} y^{5}+28 x^{2} y^{6}-8 x y^{7}+y^{8}}\end{array}$$

Exercise $$\PageIndex{7}$$

1. Determine the factorials of the integers $$5, 10, 15, 20$$, and $$25$$. Which grows faster, the common exponential function $$a_{n} = 10^{n}$$ or the factorial function $$a_{n} = n!$$? Explain.
2. Research and discuss the history of the binomial theorem.

25The product of all natural numbers less than or equal to a given natural number, denoted $$n!$$.
26The factorial of zero is defined to be equal to $$1; 0! = 1$$.
27An integer that is calculated using the formula: $$\left( \begin{array}{l}{n} \\[4pt] {k}\end{array}\right)=\frac{n !}{k !(n-k) !}$$
28Describes the algebraic expansion of binomials raised to powers: $$(x+y)^{n}=\sum_{k=0}^{n} \left( \begin{array}{l}{n} \\[4pt] {k}\end{array}\right) x^{n-k} y^{k}$$.