The Binomial Theorem

Powers of Binomials Pascal’s Triangle The Binomial Theorem Factorial Identities … and why The Binomial Theorem is a marvelous study in combinatorial patterns.

Enter nCr(4,{0,1,2,3,4}) Enter 4 nCr {0,1,2,3,4} This finds the binomial coefficients for n = 4. The calculator gives the list {1, 4, 6, 4, 1}. Using these coefficients, construct the expansion:

The triangle starts with a pyramid of 1’s. To make a new row, you add 2 entries in the previous row together and put the answer underneath and between the previous entries. You sandwich each row with 1’s

When you count rows or entries, start with 0. Row 0 Row 1 Row 2 Row 3 Row 4 Row Position If I ask for 5 C 3, I go to row 5 and choose the 3 rd entry

Pascal’s triangle can be used for combinatoric problems. These are problems where you want to see how many different ways you choose a subset from a full group. For example, choosing 3 students from a group of 5. This is fine for small problems, but isn’t so nice for big ones. How many different ways can I draw 5 cards from a deck of 52? It’s just the 5 th entry on the 52 nd row of the triangle. OR we can use a formula.

This isn’t quite the same as a poker hand, since poker only cares about suites when you look at a flush, but you’ll get the idea. If I want to see how many unique combinations of 5 cards I can make from a deck of 52, then I need 52 C 5. Tip: count the top down to the larger of the 2 numbers in the bottom, then reduce. Only count down the small one in the denominator. The 47! will cancel and you can reduce in the rest of the fraction will reduce like normal. Then just multiply whatever is left. 2,598,960 That’s a lot of different hands!

Pascal’s triangle can also be used for binomial expansion. The exponent tells you what row to go to and the row entries are the coefficients The coefficients (blue) come from the triangle. The red part is never written, but notice how the exponents count down for x and up for y.

1.Copy the coefficients for the row that matches your exponent. 2.Let the exponent on the first term count down to 0. 3.Let the exponent on the second term count up from 0. 4.If the sign is +, then they are all + 5.If the sign is -, then they alternate + and -.

1. Take the coefficients from row The (2x) starts with an exponent of 5 and counts down. 1(2x) 5 5(2x) 4 10(2x) 3 10(2x) 2 5(2x) 1 1(2x) 0 3. The (3y) starts with an exponent of 0 and counts up. 1(2x) 5 (3y) 0 5(2x) 4 (3y) 1 10(2x) 3 (3y) 2 10(2x) 2 (3y) 3 5(2x) 1 (3y) 4 1(2x) 0 (3y) 5 4. Since it’s a minus in the middle, we alternate + and – signs. 1(2x) 5 (3y) 0 - 5(2x) 4 (3y) (2x) 3 (3y) (2x) 2 (3y) 3 + 5(2x) 1 (3y) 4 - 1(2x) 0 (3y) 5 5. Now we clean it up. Remember that order of operations says exponents before multiplication.

1(32x 5 )(1)- 5(16x 4 )(3y) (8x 3 )(9y 2 ) - 10(4x 2 )(27y 3 ) + 5(2x)(81y 4 ) - 1(1)(243y 5 ) It’s not much fun, but it beats doing this: Enter expand((2x – 3y)^5, x) Be sure your variables are not already assigned a value!!!!

The only term in the expansion that we need to deal with is The coefficient is −1320.

Text pg715 Exercises # 2-26 even