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Section 8 Pascal’s Triangle

Section 8 Pascal’s Triangle. Questions about homework? Submit homework!. #1. How many ways can we choose a committee of 5 from a class of one instructor and 13 students if the committee may consist of any 5 of the 14 individuals. the committee must include the instructor.

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Section 8 Pascal’s Triangle

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  1. Section 8Pascal’s Triangle • Questions about homework? • Submit homework! MATH 106, Section 8

  2. #1 • How many ways can we choose a committee of 5 from a class of one instructor and 13 students if • the committee may consist of any 5 of the 14 individuals. • the committee must include the instructor. • the committee must not include the instructor. C(14,5) = 2002 C(13,4) = 715 Why do these two answers sum to the previous answer? C(13,5) = 1287 MATH 106, Section 8

  3. Pascal’s Formula • Pascal observed the following: • C(n, k) = C(n-1, k-1) + C(n-1, k) The number of ways to choose k items from the n items The number of ways to choose k items from the n–1 items The number of ways to choose k–1 items from the n–1 items Set aside 1 “special” item leaving n–1 items. Select k items from the n items as follows: Choose the “special” item and then choose k–1 items from the n–1 items or Do not choose the “special” item and then choose k items from the n–1 items MATH 106, Section 8

  4. Pascal’s Formula • Pascal observed the following: • C(n, k) = C(n-1, k-1) + C(n-1, k) • Let’s check it! 4, 3 C( ) + C( ) 4, 2 • C(5, 3) = • C(7, 2) = • C(6, 1) = 6 4 10 6, 2 C( ) + C( ) 6, 1 6 15 21 5, 1 C( ) + C( ) 5, 0 1 5 6 MATH 106, Section 8

  5. C(1, 1) C(1, 0) C(2, 0) C(2, 1) C(2, 2) C(3, 3) C(3, 0) C(3, 1) C(3, 2) C(4, 0) C(4, 1) C(4, 2) C(4, 3) C(4, 4) C(5, 5) C(5, 0) C(5, 1) C(5, 2) C(5, 3) C(5, 4) Pascal’s Triangle C(0, 0) How do we compute the values? Note: The first row is row 0 (that is, n is 0). MATH 106, Section 8

  6. Pascal’s Triangle C(0, 0) C(1, 1) C(1, 0) C(2, 0) C(2, 1) C(2, 2) C(3, 3) C(3, 0) C(3, 1) C(3, 2) C(4, 0) C(4, 1) C(4, 2) C(4, 3) C(4, 4) C(5, 5) C(5, 0) C(5, 1) C(5, 2) C(5, 3) C(5, 4) 1 1 1 1 1 2 3 1 1 3 6 4 1 1 4 MATH 106, Section 8

  7. #2 In the Pascal’s Triangle displayed below, add the entries up to and including those corresponding to the row for n =10. Pascal’s Triangle 1 1 1 1 1 2 3 1 1 3 6 4 1 1 4 10 1 1 10 5 5 15 6 15 1 20 1 6 35 21 21 1 35 7 1 7 MATH 106, Section 8

  8. We now have two ways to calculate C(n, k) • The formula from Section 5 • Pascal’s Triangle #3 Use Pascal’s Triangle to evaluate each of the following: C(6, 3) C(10, 4) C(7, 5) C(10, 6) C(8, 4) C(10, 5) 210 20 210 21 252 70 MATH 106, Section 8

  9. Homework Hints: In Section 8 Homework Problem #5, In Section 8 Homework Problem #9(b), In Section 8 Homework Problem #10(b), you can choose to construct the triangle from the beginning, or you can choose to copy the partially constructed triangle in Section 8 of the textbook (after verifying that it is correct) and then add the required rows to complete the triangle. you can use your calculation in part (a) together with the GOOD = ALL  BAD principle. notice that there is only one way to alternate speeches and musical selection, after which the order of the speeches and the order of the musical selections must be decided. MATH 106, Section 8

  10. Section 9Binomial Expansion MATH 106, Section 8

  11. Let’s take a look back … T H • Section 7, homework problem #6 • A coin is flipped 7 times. What percent of the times will result in each of the possibilities? • You must calculate the combinations for each number of heads • One each flip, there are two possible outcomes. • How did we figure the percentages? 7 These match the row of Pascal’s Triangle corresponding to n = MATH 106, Section 8

  12. What is “Binomial Expansion”? #1 Take an expression of the form (x + y)n and multiply it out for n = 2 and for n = 3. (x + y)2 = Before simplification, how many terms did we get? (x + y)3 = Before simplification, how many terms did we get? yx + x2 + 2xy + y2 x2 + y2 = (x + y)(x + y) = xy + 4 = 22 (x + y)(x + y)(x + y) = yyx + xyy + yxy + x3 + xyx + yxx + xxy + y3 = x3 + 3x2y + 3xy2 + y3 8 = 23 RECIPE FOR CHOOSING ONE TERM IN THE EXPANSION OF (x + y)n BEFORE SIMPLIFICATION: Choose x or y from first factor and then choose x or y from second factor and then … MATH 106, Section 8

  13. #2 Without doing any multiplication, find the following expansion: (x + y)7 = If we did the multiplication, how many terms would we get before simplification? x7 + x6y + x5y2 + x4y3 +x3y4 +x2y5 +xy6 +y7 21 7 7 35 35 21 128 = 27 RECIPE FOR CHOOSING ONE TERM IN THE EXPANSION OF (x + y)7 BEFORE SIMPLIFICATION: Choose x or y from first factor and then choose x or y from second factor and then … Observe that the coefficients in the simplified expansion of (x + y)7 match the row of Pascal’s Triangle corresponding to 7. MATH 106, Section 8

  14. x y It’s the same idea as flipping a coin 7 times… • Each time we compute a term in the expansion of (x + y)7, we choose x or y seven times in a row. • Each time flip a coin seven times in a row, 7 choices of H or T are made in a row • We can use Pascal’s triangle … (x + y)7 = x7 + 7x6y + 21x5y2 + 35x4y3 + 35x3y4 + 21x2y5 + 7xy6 + y7 Notice the symmetry! MATH 106, Section 8

  15. The Binomial Theorem (bottom of page 67 in the textbook) (x + y)n = C(n, 0)xny0 + C(n, 1)xn-1y1 + C(n, 2)xn-2y2 + C(n, 3)xn-3y3 + … + C(n, n-2)x2yn-2 + C(n, n-1)x1yn-1 + C(n, n)x0yn Because of symmetry, we could choose to write the formula in the Binomial Theorem as follows: (x + y)n=C(n, n)xny0 +C(n, n-1)xn-1y1 +C(n, n-2)xn-2y2 +C(n, n-3)xn-3y3 + … + C(n, 2)x2yn-2 + C(n, 1)x1yn-1 + C(n, 0)x0yn MATH 106, Section 8

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