NOTES ON OCT 9, 2012
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1 NOTES ON OCT 9, 0 MIKE ZABROCKI Exercises: I said that I would solve problems that people asked me about in class. I am going to put the solutions to these (the ones that people asked about so that you have a reasonable idea of my expectations of what I would like to see as justification of these problems. (7 n 0 ( n+ ( n x n ( How many ways are there making change for n = $.00 with pennies, nickels, dimes and quarters such that: (a there are an even number of nickels and no pennies? (b such that there at most 6 nickels? (c the total number of nickels and dimes is even? (d the total number of pennies, dimes and quarters is even? (* ( How many ways are there of placing n = 50 balls in 0 distinguished boxes such that: (c the first 4 boxes have at most 0 of the balls? (d the first 4 boxes have at least half of the balls? (* (3 (b Find the generating function for the sequence a, a 0, a 3, a, a 5, a 4, a 7, a 6,... in terms of the generating function A(x = n 0 a nx n. (d Given D 0 =, D = a and D n+ = ad n + bd n where a, b are unknowns. The entry sequence D n will be a polynomial in a and b. Find the coefficient of a r b s. (* I will not be able to do ( (d and ( (d here. I will post a solution at a later date. ( n+ ( n x n. I will use one fact that we derived (7 I want to give an expression for n 0 on Sept 7 notes, that if A(x = n 0 a nx n, then xk k! A(k (x = ( n n 0 k an x n. I will also use the fact that x r A(x = n 0 a nx n+r = n r a n rx n. Start with A(x = ( n+ ( x 3 = n 0 x n. Notice that since ( n = (n (n 3 then what I would like to do is decrease the exponent of the x in ( n+ x n in A(x by (to n then differentiate twice and then multiply until we have the right exponent.
2 MIKE ZABROCKI A(x 3x = ( n + x n n x (A(x 3x = ( n + x n n d ( x dx (A(x 3x = ( n + d dx (xn = ( n + n n d ( x dx (A(x 3x = ( n + (n (n 3 x n 4 n x 4 d ( x dx (A(x 3x = ( ( n + n x n = ( ( n + n n n 0 (n (n 3x n 4 x n ( (a Every way of making change for n cents using an even number of nickels and some dimes and quarters is a tuple of the form (X, Y, Z where X is some means of making change for r cents with an even number of nickels, Y is some means of making change for s cents using dimes and Z is some means of making change for n r s cents with quarters. Therefore we have that the generating function for making change with an even number of nickels, and some dimes and quarters is equal to = (g.f. for the ways of making change using an even number of nickels(g.f. for the ways of making change using dimes(g.f. for the ways of making change using quarters If I am using an even number of nickels, then I can make change for n cents in one way if and only if n is a multiple of 0. Therefore g.f. for the ways of making change using an even number of nickels = x 0 Similarly, if I am making change for n cents using dimes, then I can make change for n cents in one way if and only if n is a multiple of 0. Therefore g.f. for the ways of making change using dimes = x 0 If I am making change for n cents using quarters, then I can make change for n cents in one way if and only if n is a multiple of 5. g.f. for the ways of making change using quarters = x 5 We conclude that the g.f. for the number of ways of making change for n cents using an even number of nickels and some dimes and quarters is ( x 0 ( x 5
3 NOTES ON OCT 9, 0 3 Ideally we would like to take the coefficient of x 00 and in this case the generating function is simple enough that we can do this by hand. Since = +x 5 +x 50 +x 75 +x 00 + x 5 then ( x 0 ( x 5 = x 00 ( x ( x ( x ( x ( x 0 0 Well we know that there is no way of getting a power of x 5 or x 75 in /( x 0 because the only powers that appear are multiples of 0. Moreover we also know that /( x 0 = + x 0 + 3x 0 + 4x x x Therefore, ( x 0 ( x 5 = = 8. x 00 ( (b Every way of making change with pennies, dimes, quarters and 6 nickels can be broken down into four steps consisting of a way of making change in pennies for r cents, a way of making change in dimes for s cents, a way of making change with quarters for t cents and with at most 6 nickels for n r s t cents. By the multiplication principle of generating functions, we know that g.f. for making change for n cents with pennies, dimes, quarters and at most 6 nickels = (g.f. for making change for n cents with pennies(the ways of making change using dimes(g.f. for the ways of making change using quarters(the ways of making change using at most 6 nickels The generating functions for making change with pennies, dimes and quarters are similar to the last problem and are respectively x,,. The g.f. for making change x 0 x 5 using at most 6 nickels is slightly different. Then there is exactly one way of making change for 0, 5, 0, 5, 0, 5, 30 cents using at most 6 nickels hence the generating function is given by Therefore + x 5 + x 0 + x 5 + x 0 + x 5 + x 30 = x35 x 5 g.f. for making change for n cents with pennies, dimes, quarters and at most 6 nickels = x 35 ( x( x 5 ( x 0 ( x 5 ( (c The number of ways of making change for n cents with pennies, quarters and then an even number of nickels and dimes can be seen a way of making change for r cents using pennies, followed by s cents using quarters, followed by n r s cents using an even number of nickels and dimes. Therefore, g.f. for making change for n cents using pennies, quarters and then an even number
4 4 MIKE ZABROCKI of nickels and dimes = (g.f. for making change for n cents using pennies(g.f. for making change for n cents using quarters(g.f. for making change for n cents using even number of nickels and dimes Now there is one way of making change for n cents using pennies hence the generating function is x. There is one way of making change for n cents using quarters if and only if n is divisible by 5, hence the generating function for making change for n cents using quarters is. x 5 Now to make change for n cents with an even number of nickels and dimes is only possible if n is divisible by 5. In fact, if n is divisible by 0 then there is one way to make change for 0 cents and one more way for each 0 cents more and so the generating function for these terms is /( x 0 = + x 0 + 3x x If n 5 (mod 0, then there are no ways of making change for 5 cents with an even number of coins and for every 0 cents there is one more way so the generating function for these terms is x 5 /( x 0 = x 5 + x x x If n 0 (mod 0 then there is one way for making change for 0 cents ( nickels and one more way for each 0 cents, so the generating function for these terms is x 0 /( x 0 = x 0 + x x x If n 5 (mod 0 there is one way of making change for 5 cents (one nickel and one dime and one more way for each 0 cents after so the generating function for these terms is x 5 /( x 0 = x 5 + x x x Since every multiple of 5 is equivalent to 0, 5, 0 or 5 (mod 0 then the generating function for the number of ways of making change for n cents using an even number of nickels and dimes is equal to the sum of the generating functions for n equivalent to 0, 5, 0, or 5 (mod 0, therefore the generating function is equal to ( + x 0 + x 5 + x 5 /( x 0. We conclude that g.f. for making change for n cents using pennies, quarters and then an even number of nickels and dimes = + x 0 + x 5 + x 5 ( x( x 5 ( x 0 (3 (b We know from class Sept 7 that if A(x = n 0 a nx n, then x (A(x + A( x = a 0x + a x 3 + a 4 x 5 + a 6 x 7 + and the generating function for the odd terms (shifted is hence x (A(x A( x = a + a 3 x + a 5 x 4 + a 7 x 6 + x (A(x + A( x+ x (A(x A( x = a +a 0 x+a 3 x +a x 3 +a 5 x 4 +a 4 x 5 +a 7 x 6 +a 6 x 7 +
5 NOTES ON OCT 9, 0 5 (3(d Since D 0 = and D = a, then D(x = n 0 D nx n = + ax + n D nx n, now use the recursive definition for n, so that D(x = + ax + n (ad n + bd n x n = + axd(x + bx D(x Solving for D(x, we have D(x axd(x bx D(x = D(x( ax bx =, so then D(x = ax bx. Now you are asked what is the coefficient of a r b s in the coefficient of x n. To do this we expand as a series in x, then look at the resulting expression and take the coefficient of a r b s. ax bx = (ax + bx = (ax+bx m = (a+bx m x m = ( m a m k b k x m+k. k Now the coefficient of x n forces n = m + k, hence D n = ( n k a n k b k. k k 0 The coefficient of a r b s is equal to 0 unless k = s and n k = r (or n = r + s and if n = r + s then the coefficient is equal to ( r+s s. k 0
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