General comment 1

Please give references to the theorems and propositions that you are using. For example, "by Theorem 4.2 ...". Some identities, such as Pascal's (aka Prop. 3.18) are so well known and used so often that it is less important for these. But other formulas, such as Theorems 4.3 and 4.7 are less common and should be referenced. I have not generally marked down for not giving references -- yet -- but it can make a difference in borderline cases.

General comment 2

Sometimes a student will give one or more examples of a calculation before proving the general case. While this is no doubt useful when working out the proof, these examples should be left out of the final proof write-up -- in other words, please proceed directly to the general case -- unless you are presenting some novel or complicated construction where an example is useful to clarify the description of the construction. Some students, for some problems, only presented examples. Unfortunately, "proof by example" is not an accepted technique for this class. (Note: when I say to skip the examples, I am not referring to base cases of inductive proofs. Base cases are more than examples, they are a part of the proof. My comment applies to non-base-case examples that don't particularly help clarify the presentation.)

Regarding problem 1

The scoring was 2 points each for parts (a) and (b) and four points for part (c), for a total of 8 points for this problem. (All other problems were 5 points.) Some students misunderstood the intent of the problem statement for parts (a) and (b), not realizing that the subsets B and C respectively are fixed for these parts of the problem. Part (c) specifically asks for a combinatorial (aka bijective) proof. I only gave 2 points if only a non-combinatorial proof was provided.

Regarding problem 2 (#29)

Several students asked if this was really a combinatorics problem rather than just some very simple algebra, suspicious that more must be expected for a solution. The response is that you can view the problem as counting a set of objects in two different ways, the objects being (student-class) pairings representing "registrations" (or "seats" or "slots" or "student-hours" or something similar). Yes, sometimes it really is that simple.

Regarding problem 3 (#32)

Some students went astray by using addition where they should use multiplication. This is a basic counting principle which is essential to understand (e.g. for the midterm!). For example, when counting NE paths from (0,0) to (n,m) which pass thru (a,b), you can view this as constructing a path from (0,0) to (a,b) (for which there are (a+b \choose a) possibilities) followed by a path from (a,b) to (n,m) (for which there are (n+m-a-b \choose m-b) possibilities). To get the total number of such paths you need to MULTIPLY these quantities, not add them, since the choices in the two steps are independent of each other.

Regarding Problem 6 (#44)

Most people noted that the solution involves maximizing the multinomial coefficient (6 \choose a,b,c,d) where a+b+c+d = 6 is a weak composition of 6 (i.e. zero parts are allowed), and that since this is independent of the order of the parts, one may restrict to the cases of a weak partition of 6 (i.e. can assume the parts are in increasing or decreasing order). Most also noted that since (6 \choose a,b,c,d) = 6!/(a!b!c!d!) it suffices to minimize the denominator a!b!c!d!. For this much I awarded 3 points. However, one must give some argument that this product is minimized by the partition 2+2+1+1. One method is to simply list all of the weak partitions of 6 into four parts (there are nine of them) and compute the expression for each case and point out the answer among the list. Another is to use ad hoc arguments, for example pointing out that if any part of the partition is 3 or more then the expression must be >= 6, then inspect the remaining cases (2+2+2+0 gives 8 and 2+2+1+1 gives 4). But listing only a few of the examples without any other argument lost a couple of points. A few students did not mention multinomial coefficients at all, using instead a two-step argument based on the unimodality of the binomial coefficients (apparently covered during lecture), e.g. first expressing (x_1 + x_ + x_3 + x_4)^6 as a binomial [(x_1 + x_2) + (x_3 + x_4)]^6, noting that the largest binomial coeff. here is (6 \choose 3) which is the coeff. of (x_1+x_2)^3 * (x_3+x_4)^3, and then noting that the largest coeff. in the cubic factors is (3 \choose 2), arriving at the conclusion that the largest coeff. of the original expression must be (6 \choose 3)*(3 \choose 2)^2. This is indeed the correct value, but the reasoning behind this two step process is hardly obvious and requires further justification. This approach generally received two points. (Part of the intent of the problem is to encourage facility with multinomial coefficients.)