 ```18.314 PSET 2 ANSWER KEY, GRADING CRITERIA
I negated -1 or none for minor arithmetic errors. I didn’t give partial points for any approach
that is clearly wrong. For most of the problems, I negated half credit for showing a mapping
without any logical explanation. (Except problem 5,6)
Problem 1. Omitted
Problem 2. 2n
n −1
Summing elements in each diagonal For i ∈ [0..n − 1] :
P
n+i
2n
Using Problem 1, we get n−1
i=0 n−1 = n − 1
3n
Problem 3. 3n
n − n−1
Pn−1+k
k=i
k
i
=
n+i
i+1
Number of paths that touches y = x + 1 is equal to number of paths from origin to (n − 1, 2n + 1).
3n
Hence 3n
n − n−1
Problem 4. 14
4
Replace each variable x by form 2x0 + 1. Then the answer is number of weak composition of 10
into 5 parts.
Problem 5.
For a ∈ A. Put 2 in front of a. Sum up elements of each sequence, starting with 2 and ending
just before next 2. (Ex: 2,1,1,2,1 turns into 5,3)
Problem 6.
For a ∈ A. Put 1 in front of a. Sum up elements of each sequence, starting with 1 and ending
just before next 1. (Ex: 1,1,2,2,1 turns into 1,5,1)
Problem 7.
First part is easy. For the calculation part I gave full point as long as it is really big and started
with 2406 :)
2n n
Problem 8. 3n
n
n
n /3!
Above formula explains everything.
Problem 9.
There were 3 kinds of (full) proofs.
(1) Using hooks
(3) Negating (s − 1, s − 2, &middot; &middot; &middot; , ) out of partition of distinct numbers.
A lot of people explained the map as : Divide an even number on and on untill you get odd numbers. They get included in the second case above, and I gave full points although most of them
lacked mathematical rigor.
Date: September 28, 2007.
1
For partition of n into all distinct parts,
n = x1 + x2 &middot; &middot; &middot; + xm
xi = 2pi ai , so write n by sums of ai ’s. One can show inverse process is well-defined by using any
integer can be uniquely expressed in terms of 2-adic expansion.
Problem 10. S(n − 1, k − 1)
If you check the recurrence relation for R(n, k), it is same as recurrence relation for S(n−1, k−1).
(After putting 1 in first block, there is choice of k-1 blocks for each following elements)
Problem 11.
Use induction or the combinatorial interpretation that both sides count the number of ways to
choose a cycle.