18.314 PSET 3 ANSWER KEY, GRADING CRITERIA
October 31, 2007
General Ideas For Grading
• -1 pt. for calculation error.
• -2 ∼ -4 pts. if only one step further to the answer.
• -6 ∼ -8 pts. if get the very basic idea but no constructive progress.
• 0 pt.!!!!!! if misunderstand the whole problem.
Hints and Answers
Problem 1. 266
Hints: Principle of inclusion-exclusion.
Problem 2.
Hints:
(a)Use the fact that
Dn = n!
n
X
(−1)k
k=0
k!
(b)Consider the order of the cycle which contains n, and see how the order of
the cycle changes while removing n.
n
k
Problem 3. (a) F (n, k) =
Dn−k
(b) Dn > nDn−1, when k is even. Dn < nDn−1, when k is odd.
Hints:
(a)First select k indices to be fixed and the rest must have no fixed points.
(b)Use the fact that Dn = nDn−1 + (−1)n , or deduce the inequality from the
recursion we have in the previous problem.
1
Problem 4. E=1
Hints:
Several ways to get this answer.
• Use the fact that kF (n, k) = nF (n − 1, k − 1).
• Since the expected value is additive, we just have to compute the expected
value for every fixed point and add them up.
• Double Counting.
Problem 5. 3 · 3n − 2 · 4n
Hints: use generating function or find out the characteristic roots.
Problem 6. n · n!
Hints: use generating function or first guess the formula.
Problem 7. Ai = i!S(k, i)
1
Hints: First get the formula that
nk =
k
X
i=1
n
i!S(k, i)
i
(combinatorial argument does help), then by induction on n.
1
nN
Problem 8. (a) p(N, n) =
(b) p(N, n) = 1 − nn!N S(N, n)
Pn
k+1 n
k=1 (−1)
k
(n − k)N
Hints: (b) We have
n
1 X
k n
S(N, n) =
(−1)
(n − k)N
n!
k
k=0
Problem 9. (a) 5413 (b)
Pn
s
s=0 (−1) (n − s)!
2n−s
s
Hints: (b) Use the formula of problem 11 and consider B as the main diagonal
and first diagonal above the main diagonal. The way to put s rooks in B is a
direct bijection to consider a sequence with n+1 1’s and n 0’s which 0’s cannot
be next to each other.
1
-1 pt. without checking. -2 pts. without explanation.
2
(
Problem 10. G(n) =
2(n!)2,
if n is even;
P
k
2(n!)2 nk=0 (−1)
k! , if n is odd.
Hints: Fist place the women into the table then consider the ways men can sit.
Normally, it would be n! ways, but when n is odd, it must be derangement.
Problem 11.
Hints: Consider putting k rooks into a m × n chessboard and use
inclusion-exclusion formula to compute the number of ways such that no rooks
are in B.
Problem 12.
P
n
Hints: First get the relation that k=0 3k
F (n − 3k) −
by using induction. Then consider the expansion of
n
3k+1
∞
∞
X
x3j+1
xi X x3 j
−
)
(
F (i) )(
i!
(3j)!
(3j
+
1)!
j=0
i=0
Contact: shinnyih@mit.edu
3
F (n − 3k − 1) = 0