MATH 681 Problem Set #3
This problem set is due at the beginning of class on October 1.
1. (10 points) Prove the following combinatorial identities:
(a) (5 points) Recall that S(n, k) is equal to the number of ways to subdivide an
n-element set into knonempty parts. Produce a combinatorial argument to show
that S(n, k) = kS (n−1, k) + S(n−1, k −1).
(b) (5 points) Prove that for any m < n,Pm
k=0 n
k,m−k,n−m= 2mn
m.
2. (10 points) We know that pk(n) is the number of partitions of the number ninto
exactly knonzero parts, and it has a generating function given by P∞
n=0 pk(n)xn=
xk
(1−x)(1−x2)(1−x3)(1−x4)···(1−xk).
(a) (5 points) Prove that pk(n) = pk−1(n−1) + pk(n−k) by using a direct combi-
natorial method (e.g. bijection or alternative enumerations of the same set).
(b) (5 points) Prove that pk(n) = pk−1(n−1)+ pk(n−k) by equating the generating
function P∞
n=0 pk(n)xnto the sum P∞
n=0 pk−1(n−1)xn+P∞
n=0 pk(n−k)xn.
3. (15 points) We shall determine the number anof strings of the numbers 0, 1, or 2 of
length nwhich do not contain two consecutive zeroes.
(a) (5 points) Find a recurrence relation for an, including initial cases.
(b) (5 points) Find a closed form for an.
(c) (5 points) Find a closed form for the generating function of an(you may do this
before part (b), if desired, and use it to solve part (b)).
4. (5 points) Here Fndenotes the nth Fibonacci number with initial values 1,1, so
that F0= 1, F1= 1, and F2= 2. Prove that for non-negative mand n,Fm+n=
FmFn+Fm−1Fn−1.
5. (5 point bonus) Let f(x) be a monic polynomial of degree nwith integer coefficients
and distinct (not necessarily real) roots r1, r2, . . . , rn. Show that rk
1+rk
2+rk
3+· · · +rk
n
is an integer for any positive integer k.
On two occasions I have been asked — ”Pray, Mr. Babbage, if you put into the
machine wrong figures, will the right answers come out?” In one case a member of
the Upper, and in the other a member of the Lower House put this question. I am
not able rightly to apprehend the kind of confusion of ideas that could provoke such a
question. —Charles Babbage
Page 1 of 1 Due October 1, 2009