diff --git a/mbx/s1-2-basics.mbx b/mbx/s1-2-basics.mbx index 88283177c..bf3fbce5d 100644 --- a/mbx/s1-2-basics.mbx +++ b/mbx/s1-2-basics.mbx @@ -2024,8 +2024,8 @@ many ways can you pair up the next person who isn't already paired up?

The problem suggests that you think about how to get a list from a seating - arrangement. Could every list of n distinct people come from a seating chart? - How many lists of n distinct people are there? How many lists could we get + arrangement. Could every list of n distinct people come from a seating chart? + How many lists of n distinct people are there? How many lists could we get from a given seating chart by taking different starting places?

diff --git a/mbx/s1-3-apps.mbx b/mbx/s1-3-apps.mbx index 2092620c0..eb88f45df 100644 --- a/mbx/s1-3-apps.mbx +++ b/mbx/s1-3-apps.mbx @@ -702,8 +702,8 @@

- To prove that each function from a set S of size n to a set of size less - than n is not one-to-one, we must prove that regardless of the function f that we choose, there are always two elements, say x and y, such that f(x) = f(y). + To prove that each function from a set S of size n to a set of size less + than n is not one-to-one, we must prove that regardless of the function f that we choose, there are always two elements, say x and y, such that f(x) = f(y).

diff --git a/mbx/s2-1-examples.mbx b/mbx/s2-1-examples.mbx index 3c5a6c548..99f5b50b5 100644 --- a/mbx/s2-1-examples.mbx +++ b/mbx/s2-1-examples.mbx @@ -453,7 +453,7 @@

We didn't explicitly say to use induction here, but, especially in this context, - induction is a natural tool to try here. But we don't have a variable n to + induction is a natural tool to try here. But we don't have a variable n to induct on. That means you have to choose one. So what do you think is most useful. The number of blocks in the partition? The size of the first block of the partition? The size of the set we are partitioning? Or something @@ -747,7 +747,7 @@ induction to give a second proof that Ramsey Numbers exist.

What you do need to show is that if there are R(m - 1, n) + R(m, n - 1) -people in a room, then there are either m mutual acquaintances or n mutual +people in a room, then there are either m mutual acquaintances or n mutual strangers. As with earlier problems, it helps to start with a person and think about the number of people with whom this person is acquainted or nonacquainted. The generalized pigeonhole principle tells you something @@ -1072,7 +1072,7 @@ about these numbers.

- You have an inequality involving m and n that tells you that R(n, n) > m. Suppose you could work with that inequality in order to show that if the inequality holds, then m is bigger than something. What could you conclude about R(n, n)? + You have an inequality involving m and n that tells you that R(n, n) > m. Suppose you could work with that inequality in order to show that if the inequality holds, then m is bigger than something. What could you conclude about R(n, n)?

diff --git a/mbx/s2-2-recurrence.mbx b/mbx/s2-2-recurrence.mbx index 0eab8fa45..e46cba79c 100644 --- a/mbx/s2-2-recurrence.mbx +++ b/mbx/s2-2-recurrence.mbx @@ -17,7 +17,7 @@

- Suppose that our n-element set is N = \{a_1,a_2,\dots ,a_n\}. Then a subset of N either contains an or it doesn't. In our discussion of the Pascal recurrence, we showed that the number of k-element subsets of N that contain an is the same as the number of (k - 1)-element subsets of N - \{a_n\}. The bijection we used to prove this consists of taking an away from a set containing a_n. Thus the number of subsets of N containing a_n is the same as the number of subsets of the (n - 1)-element set N - \{a_n\}. But the subsets of N not containing an are exactly the same as the subsets of N - \{a_n\}. Thus we can partition the subsets of N into two blocks, each of which has size equal to the number of subsets of N - \{a_n\}. Therefore, by the sum principle, the number of subsets of N is twice the number of subsets of N - \{a_n\}. + Suppose that our n-element set is N = \{a_1,a_2,\dots ,a_n\}. Then a subset of N either contains a_n or it doesn't. In our discussion of the Pascal recurrence, we showed that the number of k-element subsets of N that contain a_n is the same as the number of (k - 1)-element subsets of N - \{a_n\}. The bijection we used to prove this consists of taking a_n away from a set containing a_n. Thus the number of subsets of N containing a_n is the same as the number of subsets of the (n - 1)-element set N - \{a_n\}. But the subsets of N not containing a_n are exactly the same as the subsets of N - \{a_n\}. Thus we can partition the subsets of N into two blocks, each of which has size equal to the number of subsets of N - \{a_n\}. Therefore, by the sum principle, the number of subsets of N is twice the number of subsets of N - \{a_n\}.

diff --git a/mbx/s3-1-distrib-idea.mbx b/mbx/s3-1-distrib-idea.mbx index 9344a3423..ba2795529 100644 --- a/mbx/s3-1-distrib-idea.mbx +++ b/mbx/s3-1-distrib-idea.mbx @@ -474,9 +474,7 @@

Suppose we wish to place the books in Problem (satisfying the assumptions we made there) so that each shelf gets at - least one book. Now in how many ways may we place the books? (Hint: how - can you make sure that each shelf gets at least one book before you start - the process described in Problem?) + least one book. Now in how many ways may we place the books?

@@ -703,7 +701,7 @@

- We already know how to place k distinct books onto n distinct shelves so + We already know how to place k distinct books onto n distinct shelves so that each shelf gets at least one. Suppose we replace the distinct books with identical ones. If we permute the distinct books before replacement, does that affect the final outcome? There are other ways to solve this problem. diff --git a/mbx/s3-3-partitions-int.mbx b/mbx/s3-3-partitions-int.mbx index f12b022c7..7652794e4 100644 --- a/mbx/s3-3-partitions-int.mbx +++ b/mbx/s3-3-partitions-int.mbx @@ -476,7 +476,7 @@

- How many compositions are there of k into n parts? What is the maximum number of compositions that could correspond to a given partition of k into n parts? + How many compositions are there of k into n parts? What is the maximum number of compositions that could correspond to a given partition of k into n parts?

@@ -967,7 +967,7 @@

- The number of partitions of k into n parts is equal to the number of partitions of k+\binom{n}{2} into n distinct parts. The bijection from partitions of k with n parts to partitions of k+\binom{n}{2} with n distinct parts that proves this is the one that takes a partition \lambda_n\lambda_{n-1}\cdots\lambda_1 of k with \lambda_i>\lambda_{i+1} and adds i-1 to \lambda_i to get \lambda'_i. Then \lambda' is a partition into distinct parts, and the number it partitions is k+1+2+\cdots+n-1=k+\binom{n}{2}. The proof that it is a bijection is the fact that subtracting n-i from the i\/th part of a partition of k into distinct parts yields a partition of k, because part i+j is at least j smaller than part i. + The number of partitions of k into n parts is equal to the number of partitions of k+\binom{n}{2} into n distinct parts. The bijection from partitions of k with n parts to partitions of k+\binom{n}{2} with n distinct parts that proves this is the one that takes a partition \lambda_n\lambda_{n-1}\cdots\lambda_1 of k with \lambda_i>\lambda_{i+1} and adds i-1 to \lambda_i to get \lambda'_i. Then \lambda' is a partition into distinct parts, and the number it partitions is k+1+2+\cdots+n-1=k+\binom{n}{2}. The proof that it is a bijection is the fact that subtracting n-i from the i\/th part of a partition of k into distinct parts yields a partition of k, because part i+j is at least j smaller than part i.

diff --git a/mbx/s4-2-genfn-int-parts.mbx b/mbx/s4-2-genfn-int-parts.mbx index 1b622e920..9c4027bf2 100644 --- a/mbx/s4-2-genfn-int-parts.mbx +++ b/mbx/s4-2-genfn-int-parts.mbx @@ -360,7 +360,7 @@ of it as (2i) \cdot j or if you think of it as i \cdot (2j). -

How are \qchoose{m+n}{n} and \qchoose{m+n}{n} related? Prove it. (Note this is the same as asking how \qchoose{r}{s} and \qchoose{r}{r-s} are related.) +

How are \qchoose{m+n}{n} and \qchoose{m+n}{m} related? Prove it. (Note this is the same as asking how \qchoose{r}{s} and \qchoose{r}{r-s} are related.)

diff --git a/mbx/s4-3-genfn-recurrence.mbx b/mbx/s4-3-genfn-recurrence.mbx index caaec21bd..f18f21453 100644 --- a/mbx/s4-3-genfn-recurrence.mbx +++ b/mbx/s4-3-genfn-recurrence.mbx @@ -30,7 +30,7 @@ and in the right hand side, use the fact that - \sum_{i=1}^\infty a_{i-1}x^i = x\sum_{i=1}^\infty a_ix^{i-1} + \sum_{i=1}^\infty a_{i-1}x^i = x\sum_{i=1}^\infty a_{i-1}x^{i-1} =x\sum_{j=0}^\infty a_jx^j =x\sum_{i=0}^\infty a_ix^i (where we substituted @@ -147,7 +147,7 @@ Fibonacci numbers -

The sequence of problems that follows (culminating in ) describes a number of hypotheses we might make about a fictional population of rabbits. We use the example of a rabbit population for historic reasons; our goal is a classical sequence of numbers called Fibonacci numbers. When FibonacciApparently Leanardo de Pisa was given the name Fibonacci posthumously. It is a shortening of son of Bonacci in Italian. introduced them, he did so with a fictional population of rabbits.Fibonacci numbers +

The sequence of problems that follows (culminating in ) describes a number of hypotheses we might make about a fictional population of rabbits. We use the example of a rabbit population for historic reasons; our goal is a classical sequence of numbers called Fibonacci numbers. When FibonacciApparently Leonardo da Pisa was given the name Fibonacci posthumously. It is a shortening of son of Bonacci in Italian. introduced them, he did so with a fictional population of rabbits.Fibonacci numbers

diff --git a/mbx/s5-1-sizeunion.mbx b/mbx/s5-1-sizeunion.mbx index 8bfcaec25..9a417bdbd 100644 --- a/mbx/s5-1-sizeunion.mbx +++ b/mbx/s5-1-sizeunion.mbx @@ -492,7 +492,7 @@

Frequently when we apply the principle of inclusion and exclusion, we will have a situation like that of part (d) of Problem. That is, we will have a set A and subsets A_1, A_2, \ldots, A_n and we will want the size or the probability of the set of elements in A that are not in the union. This set is known as the complement complement of the union of the A_is in A, and is denoted by A \setminus \bigcup_{i=1}^n A_i, or if A is clear from context, by \overline{\bigcup_{i=1}^n A_i}. - Give the fomula for \overline{\bigcup_{i=1}^n A_i}. The principle of inclusion and exclusion generall refers to both this formula and the one for the union. + Give the fomula for \overline{\bigcup_{i=1}^n A_i}. The principle of inclusion and exclusion generally refers to both this formula and the one for the union.