Homework 8 Solutions W11

# Homework 8 Solutions W11 - EECS 203 Homework 8 Solutions...

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EECS 203: Homework 8 Solutions Section 5.4 1. (E) 8 By binomial theorem, the coeﬃcient of the term x 8 y 9 in the expansion of ( x + y ) 17 is ( 17 8 ) = 24310 . When considering the term x 8 y 9 in the expansion of (3 x + 2 y ) 17 , let x 0 = 3 x and y 0 = 2 y , meaning that (3 x + 2 y ) 17 = ( x 0 + y 0 ) 17 . If we substitute x 0 = 3 x and y 0 = 2 y , then this term becomes ( 17 8 ) (3 x ) 8 (2 y ) 9 = ( ( 17 8 ) 3 8 2 9 ) x 8 y 9 . As a conclusion, we get the coeﬃcient of the term x 8 y 9 in the expansion of (3 x + 2 y ) 17 to be ( 17 8 ) 3 8 2 9 = 81 , 662 , 929 , 920 2. (E) 24 Because ( p k ) = p ! k !( p - k )! , the numerator of the fraction is clearly divisible by p . On the contrary, none of the factors in the denominator are divisible by p because all of them must be less than p and p is prime. As a result, since we know that any combination ( p k ) are integers, we know that ( p - 1)! must be divisible by k !( p - k )! . Therefore, ( p k ) /p = ( p - 1)! k !( p - k )! must be an integer. 3. (M) 28 a) Both the left-handed side and the right-handed side are counting the number of ways to choose 2 objects out of the total of 2 n objects. Left side: Choose 2 objects from a pile of 2 n objects ( 2 n 2 ) Right side: Suppose that there are two piles each with n objects. There are three cases of choosing two objects out of the two piles. (1) The two objects are drawn from diﬀerent piles n 2 (2) The two objects are drawn from the ﬁrst pile ( n 2 ) (3) The two objects are drawn from the second pile ( n 2 ) Therefore, the left-handed side and the right-handed side are counting the same set of objects. b) The below algebraic derivation from the right-handed side to the left-handed side proves that both sides of the equation are equivalent. ± 2 n 2 ² = (2 n )! 2!(2 n - 2)! = (2 n )(2 n - 1)(2 n - 2)! (2 n - 2)!2! = (4 n 2 - 2 n )(2 n - 2)! (2 n - 2)!2! = (4 n 2 - 4 n + 2 n )(2 n - 2)! (2 n - 2)!2! = 4( n )( n - 1)(2 n - 2)! + 2 n (2 n - 2)! (2 n - 2)!2! = 4( n !) + 2 n ( n - 2)! ( n - 2)!2! × (2 n - 2)(2 n - 3) ... ( n - 1) (2 n - 2)(2 n - 3) ... ( n - 1) = 2( n !) + (2( n !) + 2 n ( n - 2)! ( n - 2)!2! = 2( n !) + 2 n 2 ( n - 2)! ( n - 2)!2! = 2( n !) ( n - 2)!2! + n 2 = 2 ± n 2 ² + n 2

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4. (C) 32 The binomial theorem is as followed: ( x + y ) n = n X j =0 ± n j ² x n - j y j In order to prove the binomial theorem by induction, the base case and the inductive case must be proven. Let
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## This note was uploaded on 02/12/2012 for the course EECS 203 taught by Professor Yaoyunshi during the Spring '07 term at University of Michigan.

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Homework 8 Solutions W11 - EECS 203 Homework 8 Solutions...

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