Math 325K: Quiz #2 Solutions
1. Mark each of the following with the word TRUE or FALSE - do not put T or F. You do not need
to justify your answers. (1 point each)
a) Suppose A and B are both non-empty, finite sets. If A ⊂ B, then |A ∪ B| = |A| + |B|.
False. Since A ⊂ B, A ∪ B = B. Therefore, |A ∪ B| = |B| < |A| + |B| since A 6= ∅.
b) Let {an }3n=0 be a sequence such that an = a3−n for all 0 ≤ n ≤ 3. Then,
P3
i=0 (−1)
i
ai = 0.
P3
i
True.
i=0 (−1) ai = a0 − a1 + a2 − a3 . Since a0 = a3−0 = a3 and a1 = a3−1 = a2 by assumption, we have that this sum is 0.
c) Suppose |A| = 2. There is a bijection between P(A) (the power set of A) and A × A.
True. Since |A| = 2, |P(A)| = 2|A| = 4 and |A × A| = |A| · |A| = 4. For finite sets, having
the same cardinality is equivalent to there being a bijection between the sets. However, we can also
prove this explicitly. Let A = {a1 , a2 }. Then, we can define a bijection f : P(A) → A × A as follows.
f (∅) = (a1 , a1 ),
f ({a1 }) = (a1 , a2 ),
f ({a2 }) = (a2 , a1 ),
f ({a1 , a2 }) = (a2 , a2 ).
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2. i) (1.5 points) How many ways can you rearrange 123456 such that the first digit is even.
There are 360 ways to do this. There are 3 choices for the first digit (either 2, 4, or 6). Now,
there are 5 choices for the second digit (only five numbers remaining since we used one for the first
digit), then 4 choices for the third digit, then 3 choices for the fourth digit, 2 choices for the fifth digit,
and then 1 choice for the final digit. The multiplication rule applies, and we have 3·5·4·3·2·1 = 360.
ii) (1.5 points) How many ways can you rearrange 123456 such that either the first digit or the
last digit is even.
We can apply the addition rule.
# { first or last digit is even } = # { first digit is even } + # { last digit is even }
− # { first and last digits are even }.
We have that #{ first digit is even} = 360 from part i). Similarly, #{ last digit is even} = 360.
Therefore, we just want to find #{ first and last digits are even}. We can do this by choosing the
first digit to be even (3 choices), choosing the last digit to be even (there are now only 2 choices left
for this), and then 4! choices to fill in the remaining spots (4 numbers left for 4 spaces). Therefore,
the multiplication rule applies and we have 3 · 2 · 4! = 144. Therefore, the desired answer is 360 +
360 - 144 = 576.
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3. (3 points) Recall that the Fibonacci sequence Fn is defined (for n ≥ 0) by
F0 = 0, F1 = 1, and Fn = Fn−1 + Fn−2 for n ≥ 2.
Prove that F3n is even for all n ≥ 0 (i.e. show that the terms of the sequence F0 , F3 , F6 , . . . are
even). Hint: If you know F3n is even, you want to prove F3(n+1) is even, not F(3n)+1 .
Proof. We prove by induction on n.
Base Case: (n = 0) We have that F3·0 = F0 = 0 which is even. Thus, the base case is satisfied.
Induction Step: Suppose that F3n is even for some n ≥ 0. We will show that F3(n+1) is even.
This will complete the proof.
F3(n+1) = F3n+3
= F3n+2 + F3n+1
= (F3n+1 + F3n ) + F3n+1
= 2F3n+1 + F3n .
We can apply these recursion relations since 3(n + 1) ≥ 3. F3n is even by assumption. We also have
that 2F3n+1 is even (it is clearly divisible by 2). Therefore the sum, which is F3(n+1) , is even. This
completes the induction and the proof.
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