# Binomial Theorem with Combinations and not Pascal's Triangle

## Binomial Theorem - How it Works - Video

### Binomial Theorem Expanded

Binomial Theorem Expanded:

Here we have expanded our binomial (a + b). The 1st line has 1 term. The 2nd line has 2 terms. The 3rd line has 3 terms. The 4th line has 4 terms, and so on.

If we take a look at the coefficients in front of each term, the second row has 1 then 1. The 3rd row has 1, 2, 1. The 4th row has 1, 3, 3, 1. Now, instead of distributing a bunch of times, we can use Pascal's triangle to to find the coefficients in any row. With the magic of math, Pascal's Triangle is our friend.

There is a pattern with the variables as well. Can you find?

### Binomial Theorem Combination

Binomial Theorem Combination:

Instead of using Pascal's triangle to find the coefficients in front of each number. We can use combinations.

C(0, 0) = 1

C(1, 0) = 1 and C (1, 1) = 1

C(2, 0) = 1 and C (2, 1) = 2 and C(2, 2) = 1

C(3, 0) = 1 and C (3, 1) = 3 and C(3, 2) = 3 and C(3, 3) = 1 and so on.

### Properties of the Binomial Theorem

Properties of the Binomial Theorem:

The first property states what we stated already in another way. The power outside the parenthesis tells us how many terms we have after simplifying everything. In the case of (a + b)0. The 0 is n so n + 1 = 0 + 1 = 1 so there is 1 term. If we have (a + b)7. N is 7 so n + 1 = 7 + 1 = 8 so there are 8 terms.

The second property states that the first variable in the parenthesis starts with power outside the parenthesis and decreases each time, while the second variable does the opposite. Finally, when we add the two powers, the result will be same as a power at the beginning.

The third property states how it looks. There must be integers for our powers.

The forth property states another way on how to find the coefficient of the next term.

### Example 1 Part 1

Example 1 Part 1:

Here we followed the same set up that we do when using Pascal's triangle, but we use combinations instead of Pascal's triangle.

C(3, 0) = 1 and C (3, 1) = 3 and C(3, 2) = 3 and C(3, 3) = 1. Now let's substitute those numbers into the next line. If you can't remember Pascal's triangle then you can use combinations to find the coefficients.

### Example 1 Part 2

Example 1 Part 2:

For line 1 - We first start with the combinations C (3, 0), and start with x raised to the power outside the parenthesis and start with the 5 raised to the power of 0 and then moved down the line either increased the power or decreasing.

For line 2 - We simplified the combinations.

For line 3 - We simplified the exponents.

For line 4 - We simplified any number that had a power.

For line 5 - We multiplied the integers and wrote the final answer.

### Example 2 Part 1

Example 2 Part 1:

Here we followed the same set up that we do when using Pascal's triangle, but we use combinations instead of Pascal's triangle.

C(4, 0) = 1 and C (4, 1) = 4 and C(4, 2) = 6 and C(4, 3) = 4 and C(4, 4) = 1. Now let's substitute those numbers into the next line. If you can't remember Pascal's triangle then you can use combinations to find the coefficients.

### Example 2 Part 2

Example 2 Part 2:

For line 1 - We first start with the combinations C (3, 0), and start with x raised to the power outside the parenthesis and start with the 5 raised to the power of 0 and then moved down the line either increased the power or decreasing.

For line 2 - We simplified the combinations.

For line 3 - We simplified the exponents.

For line 4 - We simplified any number that had a power.

For line 5 - We multiplied the integers and wrote the final answer.