Download Exam 681: Combinatorial Identities and Paths in Grids and more Exams Mathematics in PDF only on Docsity!
- (8 of 12 students attempted this) Prove the combinatorial identity
∑n k=1 k 2 (n k
n 2 n−^1 + n(n − 1)2n−^2. You may use any method you like. A simple approach is purely combinatorial: the left side of the equation represents the number of ways to choose a subset of size k from an n-element set (which can be done in
(n k
ways)and then select, with order, an element and then another (possibly identical) element of the subset (which can be done in k^2 ways). These numbers are added up over all values of k, so that the left side represents the number of ways to choose a subset of any size, and then choose two (not necessarily distinct) elements to distinguish individually. In a less formal setting, the objects enumerated by this left side might be considered to be the number of ways to choose a k-person committee from n people, and then to select a chair and secretary (who might be the same person). In this context, the right side can be seen to enumerate the same thing through a casewise division: if the chair and secretary are the same, then we could preemptively choose the chair/secretary any of n ways, and then assign committee membership to the remaining n − 1 prospectives in any of 2n−^1 ways. On the other hand, if the chair and secretary were different, we would choose any of the n people to be the chair, any of the remaining n − 1 to be the secretary, and then committ3ee membership for the remaining n − 2 people could be resolved in any of 2n−^2 ways. There are other approaches possible, using either generating functions or direct algebra, but this is easily the simplest approach.
- (12 of 12 students attempted this) Answer the following questions relating to paths through the following grid (note the excluded portions):
(a) (5 points) How many walks are there from the lower left corner to the upper right corner taking upwards and rightwards steps only? If the grid were complete, there would be
3
routes from the lower left corner (0, 0) to the upper right (6, 3). We must remove those which pass through the excluded points (4, 2) or (5, 0) — we don’t need to exclude (6, 0) since any path through (6, 0) would have to go through (5, 0) and thus already be excluded. Exactly
2
1
paths pass through (4, 2), since there are
2
paths there and
1
paths from there to (6, 3). Likewise, exactly
0
1
paths go through (5, 0). No paths go through both points, so re-inclusion of their overlap is not necessary, and we get the result
3
2
1
0
1
(b) (5 points) How many of these walks pass through the point marked with a solid dot? If the grid were complete,
2
1
paths would go through this point. However, we must remove those paths through this point that also go through (4, 2) (it is impossible to go through both the solid dot and (5, 0), so we don’t need to
worry about that exclusion. There are
2
0
1
paths through both (2, 2) and (4, 2), so the number of paths through (2, 2) not going through (4, 2) is
2
1
4 2
0
1
= 12, which coluld actually be enumerated through brute force if you were so inclined.
- (10 of 12 students attempted this) Answer the following questions about the num- ber of solutions an to the equation x 1 + x 2 + x 3 + x 4 = n subject to the conditions that all the xi are non-negative integers, that x 1 ≤ 3 , 2 ≤ x 2 ≤ 4 , x 3 ≥ 5 , and x 4 ≥ 5.
(a) (5 points) Find a closed form for the ordinary generating function
n=0 anx
n.
The OGF for selecting x 1 is 1 + x + x^2 + x^3 , or, more compactly, 1 −x 4 1 −x. The OGF for selecting x 2 is x^2 + x^3 + x^4 , or x
(^2) −x 5 1 −x. The OGF for both^ x^3 and^ x^4 ’s selection is x^5 + x^6 + x^7 + · · · = x
5 1 −x. Thus, the OGF for selecting all of these values is either (1−x
(^4) )(x (^2) −x (^5) )x 10 (1−x)^4 or^
(1+x+x^2 +x^3 )(x^2 +x^3 +x^4 )x^10 (1−x)^2 , depending on preference — the first form is much easier to work with in part (b). (b) (5 points) Either using your generating function or by other means, find a for- mula for an. Using the generating function:
(1 − x^4 )(x^2 − x^5 )x^10 (1 − x)^4
x^12 − x^15 − x^16 − x^19 (1 − x)^4
= (x^12 − x^15 − x^16 + x^19 )
∑^ ∞
n=
n + 4 − 1 4 − 1
xn
∑^ ∞
n=
n + 3 3
xn+12^ −
∑^ ∞
n=
n + 3 3
xn+15^ −
∑^ ∞
n=
n + 3 3
xn+16^ +
∑^ ∞
n=
n + 3
∑^ ∞
n=
n − 9 3
xn^ −
∑^ ∞
n=
n − 12 3
xn^ − 2
∑^ ∞
n=
n − 13 3
xn^ +
∑^ ∞
n=
n − 16 3
∑^ ∞
n=
[(
n − 9 3
n − 12 3
n − 13 3
n − 16 3
)]
xn
so an =
(n− 9 3
(n− 12 3
(n− 13 3
(n− 16 3
This result could also be attained using inclusion-exclusion and pre-emptive as- signment of elements: pre-emptively assign a value of 2 to x 2 , and 5 to each of x 3 and x 4 , leaving n − 12 to freely distribute; then exclude those cases assigning more than 4 of the remained to x 1 , or more than 4 of the remainder to x 2 , then re-include the cases doing both.
- (6 of 12 students attempted this) Prove the combinatorial identity
∑n m=
(n m
S(n− m, k) = (k + 1)S(n, k + 1). You may use any method you like. The right side of this equation represents the number of partitions of a n-element set into k + 1 nonempty sets (which can be done S(n, k + 1) ways), and then the selection
1 + x + x 2 2! +^
x^3 3! +^ · · ·^ =^ e
x. Thus the EGF for including the appropriate nubmers of any of these letters in a string is the product
(ex^ − 1)^2 (ex)^2 = e^4 x^ − 2 e^3 x^ + e^2 x
(b) Either using your generating function or by other means, find a formula for an.
From the generating function:
e^4 x^ − 2 e^3 x^ + e^2 x^ =
∑^ ∞
n=
(4x)n n!
∑^ ∞
n=
(3x)n n!
∑^ ∞
n=
(2x)n n!
∑^ ∞
n=
(4n^ − 2 · 3 n^ + 2n)
xn n!
so an = 4n^ − 2 · 3 n^ + 2n. This result could also be obtained by inclusion-exclusion, counting all strings, excluding those omitting either A or B, and re-including those that omit both A and B.
