- Answer the problems on the exam pages.
- There are six problems for 100 total points. Actual scale is A=90, C=60.
- If you need extra space use the back of a page.
- No books, notes, calculators, or collaboration.
- The first four questions are true/false, with five points for the correct boolean answer and up to five for a correct justification.
- The parts of Question 5 have
numerical answers --
you may give your answer in the form
of an expression using arithmetic operations, powers, falling powers, or the
factorial function. If you give your answer using the "choose" notation, also
give it using only operations on this list. In addition, if your answer is a
**non-negative integer less than or equal to 100**, you must compute the number for full credit.

Q1: 10 points Q2: 10 points Q3: 10 points Q4: 10 points Q5: 30 points Q6: 30 points Total: 100 points

- The first two true-false questions involve a situation where three six-sided dice are thrown. A "throw" is an outcome such as "die 1 is 5, die 2 is 3, die 3 is 6".
**Question 1 (10):***True or false with justification:*The number of possible throws with at least one six is greater than the number of throws without any sixes.FALSE. There are 6

^{3}= 216 total throws. It's easiest to count the throws with no sixes, of which there are 5^{3}= 125, more than half of 216. You could also count the throws with sixes -- there are 36 of the form 6xx, 36 of the form x6x, and 36 of the form xx6, for a total of 108 minus some double-counting. This is enough to answer the question -- since there are fewer than 108 throws with sixes, the statement is false. To continue to get the exact count, though, you would remove 6 each for the throws of the form 66x, 6x6, and x66, and add one back for 666, for a total of 108 - 3(6) + 1 = 91. It's easier to get the 91 as 216 - 125, of course.**Question 2 (10):***True or false with justification:*The number of throws that contain at least one five*and*at least one six is exactly 6^{3}- [5^{3}+ 5^{3}- 4^{3}].TRUE. The number with both is 6

^{3}minus the number that either have no five or have no six. This latter number is the size of the union of "no five" and "no six", which by inclusion/exclusion has size 5^{3}+ 5^{3}- 4^{3}since the intersection of these two sets ("no five and no six") has size 4^{3}.You could also evaluate this expression as 30 and count the throws with both a five and a six by other means. There are six permutations each of 561, 562, 563, and 564, and three permutations each of 565 and 566, for a total of 30.

**Question 3 (10):***True or false with justification:*Assume that Zane's Noodle Stand offers 13 kinds of vegetables that may be put in your soup. Then the number of ways to choose*exactly five*different kinds of vegetables is strictly less than (13 choose 5).FALSE. The number of sets of five of the 13 vegetables is (13 choose 5), which is

*equal*to (13 choose 8), by the solution of the Third Counting Problem.**Question 4 (10):***True or false with justification:*Assume that eight horses are running a race, and that an "outcome" of the race is defined to be which horse is first, which is second, and which is third. Then there are over 300 possible outcomes of the race.TRUE. By the solution to the Second Counting Problem, the number of outcomes is 8

^{3}= 8*7*6 = 336 > 300. It might be quicker to note that 56*6 > 50*6 = 300.**Question 5 (30):**A sushi bar offers four kinds of rolls: Alaskan, California, Salmon, and Tuna. Five customers come into the bar and order one roll each. The waiter writes down which customer ordered which kind of roll, so a "waiter-order" might be "1-A, 2-T, 3-T, 4-C, 5-A". The waiter then gives the sushi chef a "chef-order", which in this case would be "two A, one C, two T".- (a,5) How many different possible waiter-orders are there from the five
customers?
First Counting Problem with n=4 (things to choose from) and k=5 (number of choices made), answer 4

^{5}= 1024. - (b,5) How many different possible chef-orders are there from the five
customers?
Fourth Counting Problem with n=4 and k=5, answer (n-1+k choose k) = (8 choose 5) = (8 choose 3) = 8*7*6/1*2*3 = 56.

- (c,5) In how many of the waiter-orders does each customer order a
different kind of roll?
You could use common sense and say that five customers cannot possibly each order a different kind when there are only four kinds, so the answer is zero. (This particular bit of common sense has the formal name of the Pigeonhole Principle.) Or you could blindly follow the formula for the Second Counting Problem with n=4 and k=5, getting 4

^{5}= 4*3*2*1*0 = 0. Either method works. - (d,5) How many possibilities are there for the
*set*of roll types ordered by the five customers? (In our example the set is {A, C, T}.)Any subset of {A,C,S,T} is possible except for the empty set -- since five rolls are ordered, at least one type of roll must be ordered. So the answer is 2

^{4}- 1 = 15. You could also count the one-element, two-element, three-element, and four-element sets separately by the solution to the Third Counting Problem, getting (4 choose 1) + (4 choose 2) + (4 choose 3) + (4 choose 4) = 4 + 6 + 4 + 1 = 15. - (e,5) How many different waiter-orders could result in the chef-order
"two A's, one C, one S, one T"?
We must pick which two customers order A, which can be done in (5 choose 2) = 10 ways. Then we must assign the three other customers to the three other kinds of roll, which can be done in 3

^{3}= 6 ways. So there are 60 possible waiter-orders. Alternatively, you could look at the 5! = 120 permutations of the five customers, and note that there are*two*of these for every waiter-order, meaning that there must be 120/2 = 60 waiter-orders. - (f,5) How many waiter-orders for the five customers result in all four
kinds of roll being ordered at least once? (Hint: Use the result of part (e).)
There are four possible chef-orders with all four kinds of roll ordered at least once: "AACST", "ACCST", "ACSST", and "ACSTT". We've just counted 60 waiter-orders for the first of these, and the others also have 60 waiter-orders each by an almost identical argument. Thus there are 4*60 = 240 waiter-orders that include all four kinds of roll.

- (a,5) How many different possible waiter-orders are there from the five
customers?
**Question 6 (30):**This problem explores two facts about the number (n choose 3).- (a,5) For n=3, n=4, and n=5, verify the identity:
(n choose 3) = "the sum from 1 to n of (i-1)(n-1)".
- n=3: 0*2 + 1*1 + 2*0 = 1, (3 choose 3) = 3*2*1/1*2*3 = 1
- n=4: 0*3 + 1*2 + 2*1 + 3*0 = 4, (4 choose 3) = 4*3*2/1*2*3 = 4
- n=5: 0*4 + 1*3 + 2*2 + 3*1 + 4*0 = 10, (5 choose 3) = 5*4*3/1*2*3 = 10

- (b,10) Prove the identity in part (a) for all non-negative integers n.
(Hint: Consider all the three-element subsets of the set {1,...,n}. Divide
them into groups based on their middle element, and count each group.)
For any n, if there are any three-element sets at all, they each have a middle element which must be in {1,...,n}. If that middle element is i, then the low element must be in the set {1,...,i-1} and the high element must be in the set {i+1,...,n}. There are i-1 element in the first set and (n-i) in the second, so by the Product Rule there are (i-1)(n-i) sets with middle element i. Since no two sets with different middle elements could be the same set, we can use the Sum Rule to get the total by adding the number with each middle element, getting the sum for i from 1 to n of (i-1)(n-i) as desired.

It's possible though more difficult to prove this identity by induction. After checking the base cases, you assume the identity for arbitrary k and look at the version for n = k+1. This says that (k+1 choose 3) should be the sum for i from 1 to k+1 of (i-1)(k+1-i). Breaking up the last factor of each product but the last into k-i and 1, we get that this sum is the sum of:

- the sum for i from 1 to k of (i-1)(k-i),
- the sum for i from 1 to k of (i-1)(1), and
- the final term for i = k+1, which is (k+1-1)(k+1-k+1) = 0

The first sum is (k choose 3) by the inductive hypothesis. The second sum, 0 + 1 + ... + k-1, is "well-known" to be (k choose 2). And by Pascal's Identity, (k+1 choose 3) is equal to (k choose 3) + (k choose 2).

- (c,15) Prove by induction that (n choose 3) = "the sum for i from 0 to
n-1 of (i choose 2)". (Hint: Remember that you need to prove the base cases
for n=0, n=1, n=2, and n=3, then prove that the case for n implies the case
for n+1. You may find Pascal's Identity useful -- this says that for any
non-negative integers m and k, (m+1 choose k) = (m choose k) + (m choose k-1).)
- Base case n=0, (0 choose 3) = 0, empty sum is 0
- Base case n=1, (1 choose 3) = 0, (0 choose 2) = 0
- Base case n=2, (2 choose 3) = 0, (0 choose 2) + (1 choose 2) = 0 + 0 = 0
- Base case n=3, (3 choose 3) = 1, (0 choose 2) + (1 choose 2) + (2 choose 2) = 0 + 0 + 1 = 1
- General case, let n be arbitrary and assume (n choose 3) = sum for i from 1 to n-1 of (i choose 2). Look at (n+1 choose 3), which by Pascal's Identity is (n choose 3) + (n choose 2). By the inductive hypothesis, this is (sum for i from 1 to n-1 of (i choose 2)) + (n choose 2), which equals (sum for i from 1 to n of (i choose 2)), which is what we want.

- (a,5) For n=3, n=4, and n=5, verify the identity:
(n choose 3) = "the sum from 1 to n of (i-1)(n-1)".

Last modified 20 February 2009