Illinois State University Mathematics Department

 MAT 409: Topics in Algebra and Combinatorics for K-8 Teachers

## Notes on Assignment #7

Chapter 3 Section 4
• #14: (a) There are 510 uncollected terms. (b) R = (10!)/(3!1!2!4!)
• #20: (w - 2k)4 = C(4,0)w4(-2k)0 + C(4,1)w3(-2k)1 + C(4,2)w2(-2k)2 + C(4,3)w1(-2k)3 + C(4,4)w0(-2k)4= w4 + 4w3(-2k) + 6w2(-2k)2 + + 4w(-2k)3 + (-2k)4 = w4 - 8w3k + 24w2k2 - 32wk3 + 16k4
• #27: (a) 413 uncollected terms; (b) C = 13!/(2!3!5!3!)
Chapter 4
• #1: We know that the number of uncollected terms is 515, so, after collecting terms, the sum of the coefficients must still be 515.
• #4: Arrange the 8 Qs. There's just one way to do this, assuming the Qs are identical. Now, among the 8 Qs, there are 9 spots for the Ms such that the Qs keep the Ms non-adjacent. Choose 4 of these spots and place the Ms. This can be done in C(9,4) ways.
• #5: We'll speak in terms of \$5 units. There are 20 \$5 units in \$100. (a) This problem is analogous to determining the number of positive integer solutions to the equation x1 + x2 + ... + x10 = 20. Because there are 10 categories, a tally sheet would have 9 dividers. Because there are 20 objects (\$5 units) to separate among the 10 categories, there are 19 places among the 20 objects into which we can place the 9 dividers. This can be done C(19,9) ways. (b) The phrase "at most" indicates one or more of the 10 charities could get nothing. This version of the problem is analogous to determining the number of non-negative integer solutions to the equation x1 + x2 + ... + x10 = 20. Because there are 10 categories, a tally sheet would have 9 dividers. There are 20 objects (\$5 units) to distribute among up to 10 categories. Thus, we have 29 elements to arrange, 9 of which are identical and the other 20 are identical. There are C(29,9) = C(29,20) arrangements.
• #8: We are determining the number of non-negative integer solutions to the equation z + y + x + a + b + c = 28. Because there are 6 variables, a tally sheet would have 5 dividers. There are 28 units to distribute among up to 6variables . Thus, we have 33 elements to arrange, 5 of which are identical and the other 28 are identical. There are C(33,5) = C(33,28) arrangements. Thus, there are C(33,5) = C(33,28) non-negative integer solutions to the equation z + y + x + a + b + c = 28. Any example solution that includes at least one variable equal to 0, and the six sum to 28, is correct. One such equation is 4 + 0 + 20 + 1 + 0 + 3 = 28.
• #9: (a) There are 15 elements in the set, 1 of type A, 2 of type B, 3 of type C, 4 of type D, and 5 of type E. Thus, there are 15!/(1!2!3!4!5!) distinct arrangements. (b) There are 6 vowels and 9 non-vowels. Use the 9 non-vowels as dividers. These 9 non-vowels can be arranged in 9!/(2!3!4!) ways. Now, among the 9 non-vowels, there are 10 spots into which we can place the vowels so that the vowels will be non adjacent. Choose 6 of these 10 spots, which can be done in C(10,6) ways. Once the spots for the vowels are selected, we determine the number of ways to arrange the vowels. This can be done in 6!/(1!5!) ways. We use the multiplication principle to determine the total number of arrangements, because we are matching the results of each step. This means there are 9!/(2!3!4!) × C(10,6) × 6!/(1!5!) ways to arrange the 15 elements of the set so that no two vowels are adjacent.
• #14: In (a + b + c)10, the coefficient 10!/(3!5!2!) means that one variable is raised to the 3rd power, another to the 5th power, and another to the 2nd power. There are 3! = 6 ways to arrange the three variables, so there must be 6 collected terms, including 10!/(3!5!2!) × a3b5c2, 10!/(3!5!2!) × a3c5b2, 10!/(3!5!2!) × b3a5c2, 10!/(3!5!2!) × b3c5a2, 10!/(3!5!2!) × c3a5b2, and 10!/(3!5!2!) × c3b5a2.
• #21: The two groups will have exactly the same number of subsets! The 30th row of Pascal's Triangle shows us the number of subsets from a 30-element set: C(30,0), C(30,1), C(30,2), . . . , C(30,29), C(30,30). There are 31 such entries in row 30, and the sum C(30,0) + C(30,2) + ... + C(30,28) + C(30,30) is the number of subsets with an even number of pliers. The sum C(30,1) + C(30,3) + ... + C(30,27) + C(30,29) is the number of subsets with an odd number of pliers. These two sums are identical, each equal to (1/2) × 230 = 229. This is true for every row in Pascal's Triangle. Can you justify why?
• #25: We can begin with the middle digit. There are 10 options for this digit. Next, choose the digit for the left-most (millions) position. This cannot be 0, so there are 9 options here. Between the middle digit and the left-most (millions) digit ther are two places for digits. Each can be filled in 10 ways. Once we have selected digits for these four positions in the number, we have no remaining choices if a palindrome is required. Thus, by the multiplication principle, there are 9 × 103= 9000 7-digit palindromes.