Actual scale was A=92, C=62. Question text is in black, solutions in blue.
Q1: 25 points Q2: 30 points Q3: 20 points plus 10 extra credit Q4: 20 points Total: 95 points plus 10 extra credit
First counting problem, count sequences of three customers, answer 53 = 125.
Fourth counting problem, count multisets of three customers, answer (3+5-1 choose 3) =
(7 choose 3) = (7*6*5)/(1*2*3) = 35.
Some people got this by adding the five ways for all three movies to go the same
customer, plus the 5*4 = 20 ways to give two movies to one and one to another, plus the
ten ways to give one each to three different customers counted in (d). This is
5 + 20 + 10 = 35, which is correct though more work than using the fourth counting
problem.
Second counting problem, count sequences of three customers with no repeated element, answer 53 = 5*4*3 = 60.
Third counting problem, count sets of three customers, answer (5 choose 3) = (5*4*3)/(1*2*3) = 10.
Each customer picks a subset of the three movies. There are 23 = 8 such subsets (also computable as (3 choose 0) + (3 choose 1) + (3 choose 2) + (3 choose 3)). So we are counting sequences of five subsets, which is the first counting problem -- answer 85 = 215 = 32768.
Length 0 has λ only, length 1 has a, b, and c, and length 2 has aa, bb, and cc.
A palindrome of length n+2 must have its first and last letters equal, so it must be of the form dud where d is a letter and u is a string of length n. Since (dud)R = duRd, we see that u = uR and u is itself a palindrome. Any choice of a letter d and a palindrome u leads to a palindrome dud, and all palindromes of length n+2 are of this form. (I took off two points for not arguing that you had covered all possible palindromes.) The number of choices of d is 3, so the number of choices of dud is three times the number of choices of u. So the number of length n+2 palindromes is exactly three times the number of length n palindromes, as desired.
Let P(n) be "the number of palindromes of length n is 3(n+1)/2".
Odd-even induction, odd case: The base case is n=1, and P(1) is true because
there are three palindromes and 3(1+1)/2 = 31 = 3. For the
inductive case, assume P(n) and prove P(n+2). P(n) says that the number of length
n palindromes is 3(n+1)/2, and (b) then tells us that the number of length
n+2 palindromes is three times this number, or 3((n+1)/2)+1. By the reminder
in the hint, this number is 3(n+3)/2 which is 3((n+2)+1)/2 as
desired. (The reminder makes it unnecessary to deal separately with odd and even n in
calculating with the exponent.) So P(n) implies P(n+2) and the entire odd case follows.
Odd-even induction, even case: The base case is n=0, and P(0) is true because there
is one palindrome of length 0 and 3(0+1)/2 = 30 = 1. The inductive
case is to assume P(n) and prove P(n+2), and this is identical to the inductive
step of the odd case above because the inductive step never used the fact that n was odd.
Strong induction: We prove two base cases P(0) and P(1) as above. Then from the
argument above that P(n) implies P(n+2), we can conclude that P(n-1) implies P(n+1).
If n is at least 1, then, the strong inductive hypthesis (that P(i) is true for all i
with i ≤ n) gives us P(n-1) and thus gives us P(n+1). This completes the strong
induction and proves P(n) for all naturals n.
The number of two-card sets is (52 choose 2) = (52*51)/(1*2) = 1326. The number of
two-card sets containing an ace and a card from {10,J,Q,K} is 4*16 = 64. So the
probability is 64/1326 = a bit under 5%.
It's possible but trickier to do this by counting sequences in both the numerator
and denominator. The number of two-card sequences from the deck is 52*51 (not 52*52,
because after the first card is chosen it cannot be chosen again). The number of
sequences adding to 21 is 4*16 for the sequences of an ace followed by a ten or face card,
plus 16*4 for sequences of a ten or face card followed by an ace. The answer is
thus 128/52*51, identical to the above. Most people who tried this method forgot to
consider both orders and got half the right answer. (Many of these people, perhaps,
didn't realize that they were counting sets on the top and sequences on the bottom of
the ratio.)
If we count sets, we again have (52 choose 2) = 1326 in the denominator. In the
numerator, we may have any two of the 16 tens and face cards, so there are (16 choose 2) =
(16*15)/(1*2) = 120 such sets and the answer is 120/1326 = 60/663 = 20/221, about 9%.
In this problem the sequence counters mostly got the right answer. The number of
two-card sequences is 52*51, and the number of two-card sequences of tens or face cards
is 16*15. The ratio (16*15)/(52*51) is the same as that above. Some people also got
this by computing the chance that the first card is worth ten, 16/52, and multiplying this
by the chance that the second card is also worth ten, which is 15/51 because by this
point one of the desired cards is out of the deck.
This is a more complicated application of the same principles. I found it easiest
to count sets, and add the numerators over the common denominator of (52 choose 2):
(2)-------(3)
| \ /
| \ /
| \ /
| \/
| /\
| / \
| / \
| / \
(1) (4)
Nearly everybody got this.
0 1 1 0
1 0 1 1
1 1 0 0
0 1 0 0
The last entry of the first row tells us that there is exactly one path of length 2,
and by inspection this is from 1 to 3 to 2.
There were two principal mistakes people made on this problem. Some didn't do the
matrix multiplication and looked for the paths by hand -- this was fine if you were right
but many people found only three of the four four-step paths. (This cost 3 of the 15
points.) A larger number of people misread the question -- they found the number
of paths of each length but didn't list the paths. (This cost 5 of the 15 points.)
2 1 1 1
1 3 1 0
1 1 2 1
1 0 1 1
Last modified 24 April 2005