1.3: Combinations: The Binomial Theorem
Permutation: linear arrangements of distinct objects (when order is relevant). Combination: selection of distinct objects (when order is not relevant).
Example We start with 52 standard playing cards. a) If we draw 3 cards in succession and without replacement, then how many ways are there? b) If we simply select 3 cards at once time (with no reference to order), then how many ways are there?
1.3: Combinations: The Binomial Theorem
Permutation: linear arrangements of distinct objects (when order is relevant). Combination: selection of distinct objects (when order is not relevant).
Example We start with 52 standard playing cards. a) If we draw 3 cards in succession and without replacement, then how many ways are there? b) If we simply select 3 cards at once time (with no reference to order), then how many ways are there?
1.3: Combinations: The Binomial Theorem
Definition, Notation and Formula If we start with n distinct objects, each selection, or combination, of r of these objects, with no reference to order, corresponds to r ! permutations of size r from the n objects. Thus the number of combinations of size r from the size n is C(n, r ) =
n! P(n, r ) = , 0 ≤ r ≤ n. r! r !(n − r )!
1.3: Combinations: The Binomial Theorem
Example (1.19) Lynn and Patti decide to buy a PowerBall ticket. To win the grand prize for PowerBall one must match five numbers selected from 1 to 49 inclusive and then must also match the powerball, an integer from 1 to 42 inclusive. Lynn selects the five numbers (between 1 and 49 inclusive). This she can do in 49 Meanwhile 5 ways (since matching does not involve order). 42 Patti selects the powerball−here there are 1 possibilities. Consequently, by the rule of product, Lynn and Patti can select thesix number for their PowerBell ticket in 49 42 5 1 = 80, 089, 128 ways.
1.3: Combinations: The Binomial Theorem Example (1.21) a) A Rydell High School, the gym teacher must select nine girls from the junior and senior classes for a volleyball team, if there are 28 juniors and 25 seniors, she can make the selection in 53 9 = 4, 431, 613, 550 ways. b) If two junior and one senior are the best spikers and must be on the team, then the rest of the team can be chosen in 50 6 = 15, 890, 700 ways. c) For a certain tournament the team must comprise four juniors and five seniors. The teacher can select the four juniors in 28 4 ways. For each of these selections she has 25 5 ways to choose the five seniors. Consequently, by the rule of product, she can select her team in 28 25 4 5 = 1, 087, 836, 750 ways for this particular tournament.
1.3: Combinations: The Binomial Theorem Example (1.21) a) A Rydell High School, the gym teacher must select nine girls from the junior and senior classes for a volleyball team, if there are 28 juniors and 25 seniors, she can make the selection in 53 9 = 4, 431, 613, 550 ways. b) If two junior and one senior are the best spikers and must be on the team, then the rest of the team can be chosen in 50 6 = 15, 890, 700 ways. c) For a certain tournament the team must comprise four juniors and five seniors. The teacher can select the four juniors in 28 4 ways. For each of these selections she has 25 5 ways to choose the five seniors. Consequently, by the rule of product, she can select her team in 28 25 4 5 = 1, 087, 836, 750 ways for this particular tournament.
1.3: Combinations: The Binomial Theorem Example (1.21) a) A Rydell High School, the gym teacher must select nine girls from the junior and senior classes for a volleyball team, if there are 28 juniors and 25 seniors, she can make the selection in 53 9 = 4, 431, 613, 550 ways. b) If two junior and one senior are the best spikers and must be on the team, then the rest of the team can be chosen in 50 6 = 15, 890, 700 ways. c) For a certain tournament the team must comprise four juniors and five seniors. The teacher can select the four juniors in 28 4 ways. For each of these selections she has 25 5 ways to choose the five seniors. Consequently, by the rule of product, she can select her team in 28 25 4 5 = 1, 087, 836, 750 ways for this particular tournament.
1.3: Combinations: The Binomial Theorem Example (1.22) The gym teacher of example 1.21 must make up four volleyball teams of nine girls each from the 36 freshman girls in her P.E. class. In how many ways can she select these four teams? Call the teams A, B, C, and D. To form team A, she can select any nine girls from 36 enrolled in 36 ways. For team B the selection process 9 27 9 yields 9 possibilities. This leaves 18 9 and 9 possible ways to select teams C and D, respectively. So by the rule of product, the four teams can be chosen in 27! 18! 9! 36 27 18 9 36! )( )( )( ) =( 9!27! 9!18! 9!9! 9!0! 9 9 9 9 36! . = = 2.145 × 1019 9!9!9!9! ways.
1.3: Combinations: The Binomial Theorem Example (1.22) The gym teacher of example 1.21 must make up four volleyball teams of nine girls each from the 36 freshman girls in her P.E. class. In how many ways can she select these four teams? Call the teams A, B, C, and D. To form team A, she can select any nine girls from 36 enrolled in 36 ways. For team B the selection process 9 27 9 yields 9 possibilities. This leaves 18 9 and 9 possible ways to select teams C and D, respectively. So by the rule of product, the four teams can be chosen in 36 27 18 9 36! 27! 18! 9! =( )( )( )( ) 9 9 9 9 9!27! 9!18! 9!9! 9!0! 36! . = = 2.145 × 1019 9!9!9!9! ways.
1.3: Combinations: The Binomial Theorem Example (1.22) The gym teacher of example 1.21 must make up four volleyball teams of nine girls each from the 36 freshman girls in her P.E. class. In how many ways can she select these four teams? Call the teams A, B, C, and D. To form team A, she can select any nine girls from 36 enrolled in 36 ways. For team B the selection process 9 27 9 yields 9 possibilities. This leaves 18 9 and 9 possible ways to select teams C and D, respectively. So by the rule of product, the four teams can be chosen in 36 27 18 9 36! 27! 18! 9! =( )( )( )( ) 9 9 9 9 9!27! 9!18! 9!9! 9!0! 36! . = = 2.145 × 1019 9!9!9!9! ways.
1.3: Combinations: The Binomial Theorem Example (1.22) The gym teacher of example 1.21 must make up four volleyball teams of nine girls each from the 36 freshman girls in her P.E. class. In how many ways can she select these four teams? Call the teams A, B, C, and D. To form team A, she can select any nine girls from 36 enrolled in 36 ways. For team B the selection process 9 27 9 yields 9 possibilities. This leaves 18 9 and 9 possible ways to select teams C and D, respectively. So by the rule of product, the four teams can be chosen in 36 27 18 9 36! 27! 18! 9! =( )( )( )( ) 9 9 9 9 9!27! 9!18! 9!9! 9!0! 36! . = = 2.145 × 1019 9!9!9!9! ways.
1.3: Combinations: The Binomial Theorem
Σ notation: Theorem (1.1 The Binomial Theorem.) If x and y are variables and n is a positive integer, then n 0 n n 1 n−1 n 2 n−2 n (x + y ) = x y + x y + x y + ··· 0 1 2 n n n n 0 X n k n−k n−1 1 + x y + x y = x y n−1 n k k=0
Here,
n k
is often referred to as a binomial coefficient.
1.3: Combinations:The Binomial Theorem
Example (1.26) a) Find the coefficient of x 5 y 2 in the expansion of (x + y)7 . b) Find the coefficient of a5 b2 in the expansion of (2a − 3b)7 .
Corollary (1.1) For each integer n > 0, a) n0 + n1 + n2 + · · · + nn = 2n , and b) n0 − n1 + n2 − · · · + (−1)n nn = 0
1.3: Combinations:The Binomial Theorem
Example (1.26) a) Find the coefficient of x 5 y 2 in the expansion of (x + y)7 . b) Find the coefficient of a5 b2 in the expansion of (2a − 3b)7 .
Corollary (1.1) For each integer n > 0, a) n0 + n1 + n2 + · · · + nn = 2n , and b) n0 − n1 + n2 − · · · + (−1)n nn = 0
1.3: Combinations: The Binomial Theorem
Theorem (1.2 Multinomial Theorem ) For positive integers n, t, the coefficient of x1n1 x2n2 x3n3 · · · xtnt in the expansion of (x1 + x2 + x3 + · · · + xt )n is n! , n1 !n2 !n3 ! · · · nt ! where each ni is an integer with 0 ≤ ni ≤ n, for all 1 ≤ i ≤ t, and n1 + n2 + n3 + · · · + nt = n
1.3: Combinations: The Binomial Theorem
Example (1.27) a) In the expansion of (x + y + z)7 , find the coefficients of x 2 y 2 z 3 and x 3 z 4 . b) What is the coefficient of a2 b3 c 2 d 5 in the expression of (a + 2b − 3c + 2d + 5)16 ?
