Math 116 - Solutions to selected problems from Homework # 3
The problems below are from Section 3.6 in the text.
21. We have to count the number of 9-permutations of the set {A, D, E , R, S}with repeti-
tion numbers 1,2,2,1,3. By Theorem 3.4.2, this is 9!
2!2!3! . The number of 8-permutations
of these 9 letters is the same number, since whenever we have an 8-permutation, we
can put the leftover letter at the end to get a 9-permutation, and every 9-permutation
is obtained in this way.
32. The multiset Scontains 12 elements (counting multiplicity), so when we form an 11-
permutation, there are 3 choices (a, b or c) of which type of element to leave out. If we
leave out an a, we then have to count the number of 11-permutations of the multiset
{2·a, 4·b, 5·c}, which is 11!
2!4!5! . If we leave out a b, we have to count the number of
11-permutations of the multiset {3·a, 3·b, 5·c}, which is 11!
3!3!5! . If we leave out a c, we
have to count the number of 11-permutations of the multiset {3·a, 4·b, 4·c}, which
is 11!
3!4!4! . Hence the answer is the sum of these three numbers 11!
3!4!5! (3 + 4 + 5) = 12!
3!4!5! .
Notice that this problem could also be solved using the idea for the second half of 21.
The answer is the same as the number of 12-permutations of S, that is 12!
3!4!5! , since any
11-permutation can be made into a 12-permutation by adding the leftover letter to its
end.
39. b) Choosing 6 sticks leaves 14, divided into 7 groups of size ≥0. If no two of the 6
sticks are consecutive, then the 5 groups in the middle have at least one stick each. If
xidenotes the number of sticks in the ith group, then we need to count the number
of solutions (x1, . . . , x7) to x1+· · · +x7= 14 with x1, x7≥0 and x2, . . . , x6≥1.
Replacing xiwith yi=xi−1 for 2 ≤i≤6, we need to count the number of solutions
in nonnegative integers to x1+y2+· · ·+y6+x7= 9, which equals 9 + 7 −1
7−1=15
6.
c) This is essentially the same as (b), except x2, . . . , x6≥2. Replacing xiwith yi=
xi−2, we have to count the number of nonnegative integer solutions to x1+y2+· · · +
y6+x7= 4, which equals 4 + 7 −1
7−1=10
6.
41. There are 3 ways to choose which child will get the orange. Suppose, without loss of
generality, that the first child gets the orange, and give an apple to each of the other
children. Then there are 10 apples to give to 3 children, and there are 10 + 3 −1
3−1=
12
2ways to do this. Hence there are 3 12
2total ways to distribute the fruit.
45. a) Since we only care about how many books are on each shelf, we can let xibe the
number of books on the ith shelf. There are 20 books, so any distribution of them gives
a solution (x1, . . . , x5) in nonnegative integers to the equation x1+x2+· · · +x5= 20.
Conversely, any such solution to this equation determines how many books are on
1
