Question text is in black, my answers in blue.

**Question 1.1, posted 7 February:**In problem 6.1.4, what is meant by a "natural"?In Chapter 1, which of course you don't have, I define a "natural" to be a "natural number", meaning a non-negative integer or a member of the set {0,1,2,3,...}. This is the set of possible

*sizes*of a finite set, so it's the most natural kind of number to use for counting problems.**Question 1.2, posted 12 February:**In the Yahtzee problem, you said something in lecture about the dice maybe being different colors. But isn't 4-1-2-3-5 the same roll as 1-4-2-3-5?No! Dice problems are different from card problems because you are counting sequences rather than sets. Remember when you compute that the chance of getting 7 by throwing 2D6 is 1/6, you count the six ways of getting 7 which are 1-6, 2-5, 3-4, 4-3, 5-2, and 6-1 -- you must count 1-6 and 6-1 as separate sequences. In this case if you were to count the number of straights, you would have to consider all possible orderings of a set of values as different sequences.

**Question 1.3, posted 12 February:**In 6.2.4, I'm having a very hard time visualizing this picture. Can you help?Think of the special case where a, b, and c are each 1/2. Then the set where x ≤ a is the left half of the cube, the set where y ≤ b is the bottom half, and the set where z ≤ c is the back half. The set where at least one of these things is true is seven-eighths of the cube -- it is everything except for the 1/2 by 1/2 by 1/2 cube in the top right front corner of the original cube.

**Question 1.4, posted 12 February:**In 6.2.5, how do I deal with the numbers 2, 3, 5, and 7 themselves?The problem asks you to calculate how many numbers are either multiples of 2, multiples of 3, multiples of 5, or multiples of 7. The numbers in this set are composite

*except for*2, 3, 5, and 7. So to get the number of primes, you find the total number of numbers not in this set, then add four.**Question 1.5, posted 12 February:**Can I do 6.4.4 by induction?Yes, though that might not be the easiest way to do it. If you let P(n) be the statement "at least one of the numbers in the set {n, n-1,..., n-k+1} is divisible by k", then you can prove P(n) for all n by induction, and this implies the desired result. Some of you know about modular arithmetic and such from 250 or elsewhere and some don't, so we'll take informal answers about divisibility here.

**Question 1.6, posted 12 February:**Do I have to use induction for all parts of 6.4.6?No, parts (a) and (b) are easier without induction, using the rule that if [x

_{1},...,x_{n}] and [y_{1},...,y_{n}] are two sequences of numbers with x_{i}≤ y_{i}for each i, then the product of x's is ≤ the product of the y's. Part (a) is an excellent easy practice induction proof, but part (b) is about as hard as part (c) if you do it by induction. (If you try it, I suggest defining k to be n/2 and then using induction on k. You'll need the fact about the number e from the statement of (c).**Question 1.7, posted Friday 12 February:**Here's my idea for 6.6.4 (c), counting the seven-card hands that contain a five-card straight. There are 12870 possible five-card straights, and for each of them we could add any two of the 47 other cards, so the answer should be 12870 times (47 choose 2), right?No, because you've double-counted in every instance where a seven-card hand contains more than on straight. You could use Inclusion/Exclusion by figuring out how many hands have two or three straights, but this is messy. Look at the problem more carefully, though -- I asked how many

*of the hands without pairs*have a straight. So you need to look at the (13 choose 7) possible rank sets for a seven-card hand*without a pair*, decide how many of these contain five consecutive numbers, and then account for the suits of the cards.

Last modified 12 February 2009