Midterm #2 answers (questions have been paraphrased)
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1. What's the probability of getting a straight dealt in order?
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A. The first card can be a 2, 3, ..., 10. That's 9 possibilities. Then each of
the five cards can be any suit (but we know what the ranks must be). 
That's another factor of 4^5. Then we divide by the number of ways
of being dealt 5 cards, which is 52 * 51 * 50 * 49 * 48.

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2. How many strings with k 1s, no 000, and beginning and ending with 0?
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A. Put in the k 1s first. Now put in 0s. At the beginning you could
have either "0" or "00" before the first 1, likewise at the end, so
that's 2*2 choices. Then in between each pairs of 1s (and there are k-1 
such gaps) you can have "00", "0", or "". That's 3^(k-1) choices.
So the total is 4 * 3^(k-1).
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3a. What's the biggest a_n could be in a slow-growing sequence?
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A. Well, a_n is bounded by a_(n-1) + n, which is bounded by 
a_(n-2) + (n-1) + n, ..., which is bounded by 1 + 2 + ... + n.
That's an okay answer, though summing it and getting [n+1 choose 2]
is much nicer.
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3b. How many such sequences are there?
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A. Each time we decide on our a_n, we choose a number between a_(n-1)
and a_(n-1) + n. That's n+1 different possibilities. So we get 3 choices,
then another 4, then another 5, ..., then another n+1. Multiplying these
all together, it's (n+1)!/2 (since we skip 2).
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4. Let b_n be the number of ternary strings that don't decrease.
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4a. What's b_1, b_2?
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A. b_1 is three (0,1,2), b_2 is six (00,01,02,11,12,22).
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4b. Find a recurrence for b_n.
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A. Break our set into those that have 2, 1, or 0 at the end. The ones
with 2 at the end correspond to those counted by b_(n-1) (by ripping
the 2 off). The ones with 1 at the end must have been all 0s then all 1s;
there are n of those (depending on where the first 1 is). There's
one with 0 at the end. So
<center> b_n = b_(n-1) + n + 1. </center>

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4c. Find a formula for b_n.
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A. This is a stars and bars: we have a_0 zeroes, a_1 ones, a_2 twos,
adding up to n. So the answer is (n+2 choose 2). 
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(If you find this first, you can answer 4b with Pascal's recurrence,
which is anyway proved by essentially the same analysis used in the 
answer written there.)