Possible Solutions to Test #3

(a) K = C(10,7) = C(10,3)
(b) 2^10

2. a(1) = 1, a(n) = 2a(n-1)
Starting with the first term 1, every term is double the previous term.

(a) 9!/(4!2!)
(b) 5!/2!
First place the As so a space exists between each one: __ A __ A __ A __ A __. Because the As are indistinguishable from one another, this can be done in only one way. To maintain non-adjacency among the consonants, there exist 5 spots for the 5 consonants. There are 5! ways to permute the consonants, but we divide by 2! to account for the two Ts that can be interchanged without changing the distinguishability of the letter set.

4. {5/2, 5, 10}
We have t(2) = 5(10)/5 = 10; t(3) = 5(10)/10 = 5; t(4) = 5(5)/10 = 5/2; t(5) = 5(5/2)/5 = 5/2; t(6) = 5(5/2)/(5/2) = 5; t(7) = 5(5)/(5/2) = 10. These last two terms are the same, repectively, as the first two terms, and because the recursive definition depends on the last two terms, we will repeat the same calculations.

5. 5 mowers
We assume Jannier displays his entire set of mowers each day. For N the number of mowers available, assumed to be distinct from one another, there are N! unique mower line-ups. For the March through May sale, Biltmore's is open approximately 80 days. The smallest N such that N! is larger than 80 is 5.

Note that if we remove the first assumption, then for N mowers in a set we must consider choosing 1 from the set, arranging it, 2 from the set, arranging them, and so on, up through choosing N from the set and arranging them. For N=4: There are 4 1-element subsets and each can be arranged in 1!=1 way; there are 6 2-element subsets and each can be arranged in 2!=2 ways; there are 4 3-element subsets and each can be arranged in 3!=6 ways; there is 1 4-element subset, and it can be arranged in 4!=24 ways. This yields 4 + 12 + 24 + 24 = 64 arrangements, short of the approximately 80 required for the Spring Season Sale. With 5 mowers, this method produced in excess of 80 unique arrangements.

6. C(31,23) = C(31,8)
We can look at this as placing 8 items, each into 1 of into 24 categories. A category can have no items in it. This results in C(8+24-1,24-1)=C(31,23) solutions.

(a) C(5,3)x^2(3y)^3 = 270x^2y^3
(b) Ordered pairs of the form (k,-k/3), k not 0, result in 0^5 = 0.

8. One way to accurately represent a daily report from Nat is to use an ordered 5-tuple (c,b,e,m,t), where each position can be either a 0 or a 1 to indicate whether Nat did or did not see a particular type of vehicle. Because there are 5 elements in a report and each element is one of two choices (0 or 1), there are 2^5 = 32 possible reports that could be submitted. No month, including last month, has 32 days, so it is possible that Nat could have submitted a different report each morning last month.

9. C(14,2) = C(14,12)
We can illustrate the problem situation by using 25 tally marks to represent the 25 summers Norton has played in the band. Each tally mark belongs in one of three columns, each column representing a specific musical instrument (trumpet, trombone, tuba). We know the trumpet column has at least two tally marks, the trombone column has at least four tally marks, and the tuba column at least seven tally marks. Therefore, there are 13 tally marks already placed. There remain 12 tally marks to place, with the possibility that a column may get no more marks. Thus, we determine the number of nonnegative-integer solutions to the equation x + y + z = 12. There are C(12+3-1,3-1) such solutions.

10. Given the description of pentagonal numbers, and the recursive relationship P(n+1) = P(n) + (3n + 1), we will use induction to show that P(n) = (n/2)(3n-1) for positive integers n.

Step 1: The result holds for n=1, for P(1) = (1/2)[3(1)-1] = 2/2 = 1, the first pentagonal number.
Step 2: Assume that for n=k, P(k) = (k/2)(3k-1).
Step 3: Show that for n=k+1, P(k+1) = [(k+1)/2][3(k+1)-1], or, equivalently, that P(k+1) = [(k+1)/2](3k+2) = (3k^2+5k+2)/2.

Using the recursive relationship provided, P(k+1) = P(k) + (3k + 1). Replace P(k) in this equation with its equivalent form, assumed in Step 2. This gives us P(k+1) = (k/2)(3k-1) + (3k+1). Expand the first term and express the second term with common denominator 2: P(k+1) = (3k^2-k)/2 + (6k+2)/2. Add numerators and express over the common denominator 2: P(k+1) = (3k^2-k+6k+2)/2 = (3k^2+5k+2)/2. This last expression is just the result we we sought.

We have shown through induction that for the given description of pentagonal numbers, and the recursive relationship P(n+1) = P(n) + (3n + 1), the result P(n) = (n/2)(3n-1) holds for positive integers n.