1. 2/16/20168-7 8-7 Pascal’s Triangle

2. Pascal’s Triangle For any given value of n, there are n+1 different calculations of combinations: 0C00C00C00C0 1C01C01C01C0 2/16/20168-7 =1 1C11C1 2C02C02C12C12C22C2 =2=1 3C03C03C13C13C23C23C33C3 =3 =1 4C04C04C14C14C24C24C34C34C44C4 =4=6=4=1 n = 0, 1 calculation n = 1, 2 calculations n = 2, 3 calculations etc…

3. Pascal’s Triangle For any given value of n, there are n+1 different calculations of combinations: 0 C 0 = 1 C 0 = 2/16/20168-7 1 1 1 C 1 = 1 2 C 0 = 2 C 1 = 2 C 2 = 1 21 3 C 0 = 3 C 1 = 3 C 2 = 3 C 3 = 1 3 3 1 4 C 0 = 4 C 1 = 4 C 2 = 4 C 3 = 4 C 4 = 1 4 6 4 1

4. Pascal’s Triangle 2/16/20168-7 1 1 1 1 21 1 3 3 1 1 4 6 4 1

5. Row 0 1 Row 1 1 1 Row 2 1 2 1 Row 3 1 3 3 1 Row 4 1 4 6 4 1 Row 5 1 5 10 10 5 1 Row 6 1 6 15 20 15 6 1 Row 7? See any patterns?

6. 2/16/20168-7 Properties of Pascal’s 1.The first and the last term are always 1. 2.The second and next to last terms in the nth row are n 3.Each row is symmetric 4.The sum of the terms in row n is: 2n2n2n2n 1 111111121121133113311464114641111111121121133113311464114641

7. 2/16/20168-7 Pascal’s Triangle The number of the row always starts with 0The number of the row always starts with 0 The number of the terms in a row is always one greater than the row numberThe number of the terms in a row is always one greater than the row number 1 11111111 121121121121 1331133113311331 14641146411464114641 R n: R 0: R 1: R 2: R 3: R 4: Recursively, every term can be found by finding the sum of the two terms diagonally above it. Explicitly, every term in any row can be found using the combinations formula: Where n is the number of the row, and r+1 is the number of the term

8. 2/16/20168-7Combinations/Pascal’s Using Pascal’s Triangle What is 4 C 2 ? What is the 6 th term of row 50 of pascal’s triangle? 1 111111121121133113311464114641111111121121133113311464114641

9. Examples: Find the first 4 terms in row 9 of Pascal’s Triangle Construct row 12 if the first 6 terms in row 11 are: 1,11,55,165,330,462