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U. of Illinois MATH 413 Final Exam – Fall 2009
Answer as many problems as you can. Each question is worth 6 points (total points is 66 ). Show your work. An answer with no explanation will receive no credit. Write your name on the top right corner of each page. No external assistance permitted, including calculators of any kind. [Total time: 3 hours]
Below are some formulas you might find helpful. This list is neither guaranteed to be complete, nor do I guarantee that each formula is actually needed on the exam.
( n k
n − 1 k
n − 1 k − 1
j≥ 0
xj^ =
1 − x
∑^ n
j= 0
xj^ =
1 − xn+^1 1 − x
(x + y)n^ =
∑^ n
k= 0
n k
xkyn−k, n ∈ N
( 1 − x)−n^ =
k≥ 0
n + k − 1 n − 1
xk, n ∈ N
|A 1 ∪A 2 ∪· · ·∪An| =
i 1
|Ai 1 |−
i 1 ,i 2
|Ai 1 ∩Ai 2 |+· · ·+(− 1 )n^
i 1 ,...,in
|Ai 1 ∩· · ·∩Ain |
|Ac 1 ∩Ac 2 ∩· · ·∩Acn| = |S|−
i 1
|Ai 1 |+
i 1 ,i 2
|Ai 1 ∩Ai 2 |+· · ·+(− 1 )n+^1
i 1 ,...,in
|Ai 1 ∩· · ·∩Ain |
1
- Recall the Fibonacci numbers fn = fn− 1 +fn− 2 for n ≥ 2 with f 0 = 0 and f 1 = 1. Prove that fn is even if and only if n is divisible by 3.
Solution: We did this one in class. See Section 7.1. The claim is true for f 0 , f 1 , f 2 (even, odd, odd). It follows in general from the Fibonacci recurrence and the facts
odd+odd=even, odd+even=odd, even+odd=odd.
- Find the number of integral solutions to
x 1 + x 2 + x 3 + x 4 = 25
subject to 4 ≤ x 1 ≤ 9 , − 3 ≤ x 2 ≤ 9 , − 1 ≤ x 3 ≤ 8 and 5 ≤ x 4 ≤ 8. You may use inclusion-exclusion or generating series methods.
Solution: Either way, you need to convert the question into “normal form”. Set
y 1 = x 1 − 4, y 2 = x 2 + 3, y 3 = x 3 + 1, y 4 = x 4 − 5.
Hence the desired number of solutions is the same as
y 1 + y 2 + y 3 + y 4 = 25 − 4 + 3 + 1 − 5 = 20,
subject to
0 ≤ y 1 ≤ 5, 0 ≤ y 2 ≤ 12, 0 ≤ y 3 ≤ 9, 0 ≤ y 4 ≤ 3.
To solve the problem by inclusion-exclusion, let Ai be the set of solu- tions where yi is strictly bigger than its upper bound:
A 1 = {(y 1 , y 2 , y 3 , y 4 ) : y 1 + y 2 + y 3 + y 4 = 20, y 1 ≥ 6, y 2 , y 3 , y 4 ≥ 0 },
etc. If we only have the condition that yi ≥ 0 , then we know the number of solutions is (^) ( 20 + 4 − 1 20
We have
|A 1 | =
, |A 2 | =
|A 3 | =
, |A 4 | =
|A 1 ∩ A 2 | =
, |A 1 ∩ A 3 | =
|A 1 ∩ A 4 | =
, |A 2 ∩ A 3 | = 0
|A 2 ∩ A 4 | =
, |A 3 ∩ A 4 | =
|A 1 ∩ A 3 ∩ A 4 | = 1
and all other intersections are of size 0. Hence the desired answer is ( 20 + 4 − 1 20
−|A 1 |−|A 2 |−|A 3 |+|A 1 ∩A 2 |+· · ·+|A 3 ∩A 4 |−|A 1 ∩A 3 ∩A 4 |.
To do it by generating series, one wants
[x^20 ]( 1 +x+x^2 +· · ·+x^5 )( 1 +x+· · ·+x^12 )( 1 +x+x^2 +· · ·+x^9 )( 1 +x+x^2 +x^3 )
which one can simplify and apply the negative binomial theorem.
5.Prove that of any five points chosen within a square of side length 2 , there are two whose distance apart is at most
Brief solution: This was assigned as homework (from Section 3.4). Split the square into four 1 × 1 subsquares. By the Pigeonhole prin- ciple, at least one of the subsquares has two dots. But two dots in a 1 × 1 square can be at most a diagonal part, i.e.,
2 distance.
- Suppose a network has 15 computers and 10 printers. Every five minutes, some subset of the computers requests printers. Use the pi- geonhole principle to show that you need at least 60 “printer↔computer” connections to guarantee that if 10 (or fewer) computers want a printer, there will always be connections to permit each of these computers to use a different printer.
Solution: Suppose there were fewer than 60 printer↔computer con- nections. Then by the pigeonhole principle, at least one printer P that is connected to 5 or fewer computers. If it happened that at some mo- ment, none of these 5 computers needs a printer, but the other 10 do, we would have 10 computers accessing only 9 printers, and the desired goal could not be satisfied.
- Give the exponential generating series for the number of permuta- tions with three cycles, each of size ≤ 6 , and all other cycles of size ≥ 10. Hint: you may want to modify the fact that the generating series for cycles of size k is
k≥ 1 x
k/k.
Solution: The exponential generating series for cycles of size at most 6 is f(x) = x + x^2 /x + x^3 /x + x^4 /4 + x^5 /5 + x^6 /6.
The exponential generating series for cycles of sizre at least 10 is
g(x) = x^10 /10 + x^11 /11 + · · ·.
Consider the case of k ordered cycles k ≥ 3 where the first three are of size at most 6 and the last k − 3 are of size at least 10. By the multiplication principle for exponential generating series, the number of such ordered cycles is
f(x)^3 × g(x)k−^3.
However, permutations are unordered collections of cycles, and so the exponential g.s. for the number of permutations with this cycle struc- ture is actually f(x)^3 × g(x)k−^3 C(k, 3)
Thus applying the addition principle, summing over all k ≥ 3 gives the final answer as ∑
k≥ 3
f(x)^3 × g(x)k−^3 C(k, 3)
- In class we proved that ( n + 1 k + 1
k
k
n − 1 k
n k
Use this to solve the following problem (an in class exercise): find integers a, b, c such that
m^3 = a
m 3
m 2
m 1
and use this to find a formula for 13 + 23 + 33 + · · · + n^3.
Solution: Setting m = 1 implies c = 1. Setting m = 2 implies 8 = b+ 2 or b = 6. Finally, setting m = 3 gives 27 = a + 3b + 3c = a + 18 + 3 or a = 6. Hence,
m^3 = 6
m 3
m 2
m 1
Computing the first few cases gives
13 = 6
Summing up along columns, the LHS gives 13 + 23 + 33 + 43 + · · · + n^3. On the RHS, the first column is the case k = 3 of the given identity; the second column is the case k = 2 and the third column is the case k = 1. Hence
13 + 23 + · · · + n^3 = 6
n + 1 4
n + 1 3
n + 1 2
- Consider partitions with odd, distinct parts, such as (7, 3, 1). Next consider partitions that are self-conjugate, that is, transposing their Ferrers shapes gives the same shape. For example (4, 3, 3, 1) has Ferrers shape X X X X X X X X X X X
Call the set of self-conjugate partitions SC. This final problem ask you to prove that the number of partitions of size n with odd distinct parts equals the number of self conjugate partitions of size n. Hint: consider the above example of a self conjugate partition. “Peel” the shape into three layers of “Γ ”. This gives
X X X X X X X
X X
X ,^ X
Now straighten each layer to obtain rows of size 7 , 3 and 1 respectively, which is the partition with odd distinct parts we started with. Now generalize.
Solution: The proof follows the hint. Let On be the set of partitions of n with odd, distinct parts and let Sn be the set of partitions of n that are self conjugate. Given λ ∈ S we strip layers as suggested. First note, that since λ is self conjugate, each layer is odd in length, since it consists of a box b on the main diagonal plus the same number of boxes to the right and below b. Also, note that each layer is clearly smaller than the one preceding it. Hence after straightening the layers and stacking them, we obtain a shape with odd distinct parts. This process is clearly reversible. Given a partition with odd distinct parts, take the first part and “fold it” into a Γ shape, and repeat with the second, third etc row. Then attach these Γ -layers one inside another, giving a self conjugate partitions. That these two processes are injective is clear, and moreover, they are mutually inverse (one process simply undoes what the other process does). Hence the claim follows.
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