<meta http-equiv="refresh" content="1; url=/nojavascript/"> Binomial Coefficients ( Read ) | Probability | CK-12 Foundation
You are viewing an older version of this Concept. Go to the latest version.

# Binomial Coefficients

%
Best Score
Best Score
%
Using the Binomial Theorem
0  0  0

Solve the puzzle: I am the fifth term in the expansion of $(2x + 1)^7$ . What am I?

### Guidance

Using Pascal’s Triangle and the patterns within it are only one way to expand binomials. The Binomial Theorem can also be used to expand binomials and is sometimes more efficient, particularly for higher degree binomials. The Binomial Theorem is given by:

$(a+b)^n=\dbinom{n}{0} a^n b^0 + \dbinom{n}{1} a^{n-1} b^1 + \dbinom{n}{2} a^{n-2} b^2 + \ldots + \dbinom{n}{n-1} a^1 b^{n-1} + \dbinom{n}{n} a^0 b^n$

It can be seen in this rule that the powers of $a$ and $b$ decrease and increase, respectively, as we observed in the previous concept. Recall that the notation $\dbinom{n}{r}$ refers to the calculation of the number of combinations of $r$ elements selected from a set of $n$ elements and that $\dbinom{n}{r}={_nC}_r = \frac{n!}{r!(n-r)!}$ .

As it turns out, $\dbinom{n}{0}, \dbinom{n}{1}, \dbinom{n}{2}, \dbinom{n}{3} \ldots \dbinom{n}{n-1}, \dbinom{n}{n},$ are the elements in the $(n+1)^{st}$ row of Pascal’s Triangle. If we let $n=5$ , then we can find the coefficients as follows:

$\dbinom{5}{0}=\frac{5!}{0!5!}=1; \ \dbinom{5}{1}=\frac{5!}{1!4!}=5; \ \dbinom{5}{2}=\frac{5!}{2!3!}=10; \ \dbinom{5}{3}=\frac{5!}{3!2!}=10; \ \dbinom{5}{4}=\frac{5!}{4!1!}=5; \ \dbinom{5}{5}=\frac{5!}{5!0!}=1$

These are the elements of the $6^{th}$ row of Pascal’s Triangle: $1 \ \ 5 \ \ 10 \ \ 10 \ \ 5 \ \ 1$

The Binomial Theorem allows us to determine the coefficients of the terms in the expansion without having to extend the triangle to the appropriate row.

#### Example A

Use the Binomial Theorem to expand $(x+2y)^6$

Solution: First, in this example, $a=x$ , $b=2y$ and $n=6$ . Now we can substitute into the rule.

$(x+2y)^6=\dbinom{6}{0}x^6(2y)^0+\dbinom{6}{1}x^5(2y)^1+\dbinom{6}{2}x^4(2y)^2+\dbinom{6}{3}x^3(2y)^3+\dbinom{6}{4}x^2(2y)^4+\dbinom{6}{5}=x^1(2y)^5+\dbinom{6}{6}x^0(2y)^6$

Now we can simplify:

$&=(1)x^6(1)+(6)x^5(2y)+(15)x^4(4y^2)+(20)x^3(8y^3)+(15)x^2(16y^4)+(6)x(32y^5)+(1)(1)(64y^6) \\&=x^6+12x^5y+30x^4y^2+160x^3y^3+240x^2y^4+192xy^5+64y^6$

### More Guidance

What if we just want to find a single term in the expansion? We can use the following rule to represent the $(r+1)^{st}$ term in the expansion: $\dbinom{n}{r}a^{n-r}b^r$ . The rule is for the $(r+1)^{st}$ term because if we want the $1^{st}$ term, then $r=0$ (refer back to the Binomial Theorem expansion rule). The value of $r$ in the expansion is always one less than the term number.

#### Example B

Find the $4^{th}$ term in the expansion of $(3x-5)^8$ .

Solution: Since we want the $4^{th}$ term, $r=3$ . Now we can set up the formula with $a=3x$ , $b=-5$ , $n=8$ and $r=3$ and evaluate:

$\dbinom{8}{3}(3x)^{8-3}(-5)^3=(56)(243x^5)(-125)=-1,701,000x^5$

#### Example C

Find the coefficient of the term containing $y^5$ in the expansion of $(4+y)^9$ .

Solution:

This time, think about the rule, $\dbinom{n}{r}a^{n-r}b^r$ , and that we know that $b^r=y^5$ and thus $r=5$ . We also know that $n=9$ and $a=4$ . Now we can fill in the rest of the rule:

$\dbinom{9}{5}(4)^{9-5} y^5=(126)(256)y^5=32,256 y^5$

Intro Problem Revisit Since we want the $5^{th}$ term, $r=4$ . Now we can set up the formula with $a=2x$ , $b=1$ , $n=7$ and $r=4$ and evaluate:

$\dbinom{7}{4}(2x)^{7-4}(1)^4=(35)(8x^3)(1)=280x^3$

### Guided Practice

1. Use the Binomial Theorem to show that $(a+b)^2=a^2+2ab+b^2$ .

2. Find the coefficient of the $5^{th}$ term in the expansion of $(1-3x)^{10}$ .

3. Find the constant term in the expansion of $\left(4x^3+\frac{1}{x} \right)^4$ .

1. $(a+b)^2 &=\dbinom{2}{0}a^2b^0+\dbinom{2}{1}a^1b^1+\dbinom{2}{2}a^0b^2 \\&=(1)a^2(1)+(2)ab+(1)(1)b^2 \\&=a^2+2ab+b^2$

2. $r=4$ in the $5^{th}$ term so, $\dbinom{10}{4}(1)^{10-4}(-3x)^4=210(1)(81x^4)=17,010x^4$ . Since only the coefficient is required, we can drop the variable for the final answer: 17,010.

3. The constant term occurs when the power of $x$ is zero. Let $r$ remain unknown for the time being: $\dbinom{4}{r}(4x^3)^{4-r}\left(\frac{1}{x}\right)^r$ . Now isolate the variables to determine when the power of $x$ will be zero as shown:

We can set the variable portion of the expanded term rule equal to $x^0$ .

Then simplify using the rules of exponents on the left hand side of the equation until we have like bases, $x$ , on both sides and can drop the bases to set the exponents equal to each other and solve for $r$ .

$(x^3)^{4-r} (x^{-1})^r &=x^0 \\x^{12-3r} \cdot x^{-r} &=x^0 \\x^{12-3r-r} &=x^0 \\x^{12-4r} &=x^0 \\12-4r &=0 \\12 &=4r \\r &=3$

Now, plug the value of $r$ into the rule to get the constant term in the expansion.

$\dbinom{4}{3}(4x^3)^{4-3}\left(\frac{1}{x}\right)^3=4(4x^3)\left(\frac{1}{x}\right)^3=16.$

### Practice

Expand the following binomials using the Binomial Theorem.

1. $(x-a)^7$
2. $(2a+3)^4$

Find the $n^{th}$ term in the expansions of the following binomials.

1. $(7x-2)^5; \ n = 4$
2. $\left(6x+\frac{1}{2} \right)^7; \ n = 3$
3. $(5-a)^9; \ n = 7$
4. $\left(\frac{2}{3}x+9y \right)^6; \ n = 4$
5. Find the term with $\ x^5$ in the expansion of $(3x-2)^7$ .
6. Find the term with $\ y^6$ in the expansion of $(5-y)^8$ .
7. Find the term with $\ a^3$ in the expansion of $(2a-b)^{10}$ .
8. Find the term with $\ x^4$ in the expansion of $(8-3x)^5$ .
9. Find the constant term in the expansion of $\left(x^2+\frac{3}{x} \right)^6$ .
10. Find the constant term in the expansion of $\left(\frac{5}{x^3}-x \right)^8$ .