2.02 The binomial theorem

In Math 2, we found combinations by using the following formula:

This formula uses factorials.

Exploration

Calculate the following values:

• {}_4C_0
• {}_4C_1
• {}_4C_2
• {}_4C_3
• {}_4C_4

Compare your answers to the triangle below.

1. What do you notice?
2. What do you wonder?

The triangle we will explore today was discovered by French mathematician Blaise Pascal in the 17th century. Although we call it Pascal's triangle, students from other areas of the world know it by a different name.

• In India, it is called the Staircase to Mount Meru and attributed to Indian mathematician Halayuda from the 10th century
• In Iran, it is called Khayyam's Triangle and attributed to Arabic mathematician Al-Karaji and Omar Khayyam around the year 1000
• In China, it is called Yanghui's triangle and attributed to Chinese mathematicians Jia Xian and Yanghui in the 11th century

Back in those days, there was often no means of sharing something published in one country with other countries. Because of this, mathematicians all over the world often discovered concepts around the same time and independent of each other. There are other names for it in other countries, but we call it Pascal's triangle.

To create the triangle, we begin with 1. This is called row 0.

In the next row, we begin and end with 1. This is row 1.

For the next row, we begin and end with 1, but we find the middle number by adding the two numbers above it. This is row 2.

This pattern continues to create the triangle. We always begin with 1 and end with 1. To get the number in the middle of the row, add the two numbers above it.

The row numbering begins from 0. The row number is always the same as the second element in the row.

Although there are many interesting mathematical phenomenons found within the triangle, the one we will focus on today is it's relationship to combination. In general, we can find the values of {}_n C_r in row number n (starting at row 0) and the r value is the element in the row, (also starting at 0).

So, the value of {}_4C_2 is found in row 4 (remember to begin counting from 0) and is in position 2 (count from left to right starting from 0). We see {}_4C_2=6 and can confirm this using the combination formula:

The notation \binom{n}{r} may also be used for combinations.

Examples

Example 1

Find row 10 of Pascal's triangle.

Worked Solution

Create a strategy

In the exploration, we were given the triangle up to row 5. Now, we can continue adding the elements in the previous rows and continue until we reach row 10.

Apply the idea

Row 5 of Pascal's triangle is

The next row will begin and end with 1, and we will add the elements from row 5 to find the middle values.

Continue beginning and ending rows with 1 and adding the elements from the previous rows.

The 10th row of Pascal's triangle is 1,10,45,120,210,252,210,120,45,10,1.

Reflect and check

Keep this triangle close by so you can refer to it when calculating combinations when n\leq10.

Watch question walkthrough

Example 2

Evaluate each expression using the given approach.

a

Evaluate \binom{6}{3} using Pascal's triangle.

Worked Solution

Create a strategy

In Pascal's triangle, \binom{n}{r} is the entry in position r in row n. We start counting the rows and entries from 0.

Apply the idea

We have n=6 and r=3. The entries of Pascal's triangle up to row 7 are as follows:

Row 6 is 1,\,6,\, 15,\, 20,\, 15,\, 6,\,1. Since the count for r begins at 0, \binom{6}{3} represents the 4th entry in the 6th row. So, \binom{6}{3}=20.

Reflect and check

We can check our answer by using the combination formula with n=6 and r=3.

\displaystyle \binom{n}{r}	\displaystyle =	\displaystyle \dfrac{n!}{(n-r)!r!}	State the combination formula
\displaystyle \binom{6}{3}	\displaystyle =	\displaystyle \dfrac{6!}{(6-3)!3!}	Substitute n=6 and r=3
	\displaystyle =	\displaystyle \dfrac{6!}{3!3!}	Evaluate the difference
	\displaystyle =	\displaystyle 20	Evaluate the factorials

Watch question walkthrough

b

Evaluate {}_{7}C_{4} using the combination formula.

Worked Solution

Create a strategy

The formula is {}_{n}C_{r}=\dfrac{n!}{(n-r)!r!}.

Apply the idea

We have n=7 and r=4.

\displaystyle {}_{n}C_{r}	\displaystyle =	\displaystyle \dfrac{n!}{(n-r)!r!}	State the combination formula
\displaystyle {}_{7}C_{4}	\displaystyle =	\displaystyle \dfrac{7!}{(7-4)!4!}	Substitute n=7 and r=4
	\displaystyle =	\displaystyle \dfrac{7!}{3!4!}	Evaluate the difference
	\displaystyle =	\displaystyle \dfrac{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{(3\cdot2\cdot1)(4\cdot3\cdot2\cdot1)}	Write out the expansion of the factorials
	\displaystyle =	\displaystyle \dfrac{7\cdot5}{1}	Simplify the common factors
	\displaystyle =	\displaystyle 35	Evaluate the product and quotient

Reflect and check

Finding row 7 of Pascal's triangle would be time consuming. We can also check this using a calculator by using the button marked with {}_{n}C_{r} or \text{nCr} which may require the use of the "shift" button. The calculator should confirm that {}_{7}C_{4}=35.

Watch question walkthrough

To expand binomials in the form (a+b)^n, we previously used the distributive property to multiply {(a+b)} to itself n times. However, there are patterns in the expansions, and these patterns can be summarized into the binomial theorem.

In general, the binomial theorem says:

The symbol \sum is captial sigma, and we call it a summation. It tells us to begin by substituting the lower limit (r=0 for the binomial theorem) into the expression on the right side. This corresponds to the first term in the expansion. Then, we substitute r=1 for the second term, r=2 for the third term, and continue this pattern until we substitute r=n, the upper limit, for the (n+1)th term. Because it is a sum, we add each of the terms together.

\begin{aligned}\left(a+b\right)^n&=\sum_{r=0}^{n}\binom{n}{r}\left(a\right)^{n-r}\left(b\right)^r\\&=\binom{n}{0}a^nb^0+\binom{n}{1}a^{n-1}b^1+\ldots +\binom{n}{n-1}a^1b^{n-1}+\binom{n}{n}a^0b^n\end{aligned}

For example: \begin{aligned}(a+b)^4&=\sum_{r=0}^4\binom{4}{r}\cdot a^{4-r}\cdot b^r\\&=\binom{4}{0}\cdot a^{4-0} \cdot b^0 +\binom{4}{1}\cdot a^{4-1} \cdot b^1+\binom{4}{2}\cdot a^{4-2} \cdot b^2+\binom{4}{3}\cdot a^{4-3} \cdot b^3 +\binom{4}{4}\cdot a^{4-4} \cdot b^4\\&=a^4+4a^3b+6a^2b^2+4ab^3+b^4\end{aligned}

Looking at the exponents on the variables in an expansion, the exponents of the first term are integers that descend beginning from n and ending at 0. The exponents of the second term are integers that ascend beginning from 0 and ending at n.

Because the coefficients in the binomial theorem are combinations, the elements of Pascal's triangle can be used to evaluate the coefficients as row n will given the coefficients of the terms of \left(a+b\right)^n in order.

We can use the binomial theorem to find a specific term in the expansion without having to fully expand the binomial. Each term in the expansion of (a+b)^n is of the form: \binom{n}{r}\cdot a^{n-r}\cdot b^r where r corresponds to the (r+1)th term since we begin couting the terms from 0.

So if we need to find the 10th term, we let r=9.

When fully expanded, each expansion will have (n+1) terms because we begin counting the terms from 0. So if we need to find the second to last term and n=12, there will be 13 total terms and the second to last term will be the 12th term. In this case, r=11.

Examples