Fix minor issues (#137)

* Fix typos from #136

* Fix variables not in math mode in some hints.

* Remove duplicate hint from prob 123 (issue 129)

* Fix typo in Problem 234

* Fix missing math mode

mbx/s1-2-basics.mbx

@@ -2024,8 +2024,8 @@ many ways can you pair up the next person who isn't already paired up?
       <hint>
           <p>
         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 <m>n</m> distinct people come from a seating chart?
+        How many lists of <m>n</m> distinct people are there? How many lists could we get
         from a given seating chart by taking different starting places?
           </p>
       </hint>

mbx/s1-3-apps.mbx

@@ -702,8 +702,8 @@
       </statement>
       <hint>
         <p>
-      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 <m>f</m> that we choose, there are always two elements, say <m>x</m> and <m>y</m>, such that <m>f(x) = f(y)</m>.
+      To prove that each function from a set <m>S</m> of size <m>n</m> to a set of size less
+      than <m>n</m> is not one-to-one, we must prove that regardless of the function <m>f</m> that we choose, there are always two elements, say <m>x</m> and <m>y</m>, such that <m>f(x) = f(y)</m>.
         </p>
       </hint>
 

mbx/s2-1-examples.mbx

@@ -453,7 +453,7 @@
       <hint>
         <p>
           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 <m>n</m> 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.
       <hint>
         <p>
           What you do need to show is that if there are <m>R(m - 1, n) + R(m, n - 1)</m>
-people in a room, then there are either m mutual acquaintances or n mutual
+people in a room, then there are either <m>m</m> mutual acquaintances or <m>n</m> 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.
         </statement>
         <hint>
           <p>
-            You have an inequality involving m and n that tells you that <m>R(n, n) > m</m>. Suppose you could work with that inequality in order to show that if the inequality holds, then <m>m</m> is bigger than something. What could you conclude about <m>R(n, n)</m>?
+            You have an inequality involving <m>m</m> and <m>n</m> that tells you that <m>R(n, n) > m</m>. Suppose you could work with that inequality in order to show that if the inequality holds, then <m>m</m> is bigger than something. What could you conclude about <m>R(n, n)</m>?
           </p>
         </hint>
         <solution>

mbx/s2-2-recurrence.mbx

@@ -17,7 +17,7 @@
       </hint>
       <solution>
           <p>
-              Suppose that our <m>n</m>-element set is <m>N = \{a_1,a_2,\dots ,a_n\}</m>. Then a subset of <m>N</m> either contains an or it doesn't. In our discussion of the Pascal recurrence, we showed that the number of <m>k</m>-element subsets of N that contain an is the same as the number of <m>(k - 1)</m>-element subsets of <m>N - \{a_n\}</m>. The bijection we used to prove this consists of taking an away from a set containing <m>a_n</m>. Thus the number of subsets of <m>N</m> containing <m>a_n</m> is the same as the number of subsets of the <m>(n - 1)</m>-element set <m>N - \{a_n\}</m>. But the subsets of <m>N</m> not containing an are exactly the same as the subsets of <m>N - \{a_n\}</m>. Thus we can partition the subsets of <m>N</m> into two blocks, each of which has size equal to the number of subsets of <m>N - \{a_n\}</m>. Therefore, by the sum principle, the number of subsets of <m>N</m> is twice the number of subsets of <m>N - \{a_n\}</m>.
+              Suppose that our <m>n</m>-element set is <m>N = \{a_1,a_2,\dots ,a_n\}</m>. Then a subset of <m>N</m> either contains <m>a_n</m> or it doesn't. In our discussion of the Pascal recurrence, we showed that the number of <m>k</m>-element subsets of <m>N</m> that contain <m>a_n</m> is the same as the number of <m>(k - 1)</m>-element subsets of <m>N - \{a_n\}</m>. The bijection we used to prove this consists of taking <m>a_n</m> away from a set containing <m>a_n</m>. Thus the number of subsets of <m>N</m> containing <m>a_n</m> is the same as the number of subsets of the <m>(n - 1)</m>-element set <m>N - \{a_n\}</m>. But the subsets of <m>N</m> not containing <m>a_n</m> are exactly the same as the subsets of <m>N - \{a_n\}</m>. Thus we can partition the subsets of <m>N</m> into two blocks, each of which has size equal to the number of subsets of <m>N - \{a_n\}</m>. Therefore, by the sum principle, the number of subsets of <m>N</m> is twice the number of subsets of <m>N - \{a_n\}</m>.
           </p>
       </solution>
     </activity>

mbx/s3-1-distrib-idea.mbx

@@ -474,9 +474,7 @@
           <p>
             Suppose we wish to place the books in <xref ref="bookcase">Problem</xref>
             (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 <xref ref="bookcase">Problem</xref>?)
+            least one book. Now in how many ways may we place the books? 
           </p>
         </statement>
         <hint>
@@ -703,7 +701,7 @@
       </statement>
       <hint>
         <p>
-      We already know how to place k distinct books onto n distinct shelves so
+      We already know how to place <m>k</m> distinct books onto <m>n</m> 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.

mbx/s3-3-partitions-int.mbx

@@ -476,7 +476,7 @@
         </statement>
         <hint>
           <p>
-            How many compositions are there of <m>k</m> into n parts? What is the maximum number of compositions that could correspond to a given partition of <m>k</m> into <m>n</m> parts?
+            How many compositions are there of <m>k</m> into <m>n</m> parts? What is the maximum number of compositions that could correspond to a given partition of <m>k</m> into <m>n</m> parts?
           </p>
         </hint>
         <solution>
@@ -967,7 +967,7 @@
       </hint>
       <solution>
         <p>
-          The number of partitions of <m>k</m> into <m>n</m> parts is equal to the number of partitions of <m>k+\binom{n}{2}</m> into n distinct parts.  The bijection from partitions of <m>k</m> with <m>n</m> parts to partitions of <m>k+\binom{n}{2}</m> with <m>n</m> distinct parts that proves this is the one that takes a partition <m>\lambda_n\lambda_{n-1}\cdots\lambda_1</m> of <m>k</m> with <m>\lambda_i>\lambda_{i+1}</m> and adds <m>i-1</m> to <m>\lambda_i</m> to get <m>\lambda'_i</m>. Then <m>\lambda'</m> is a partition into distinct parts, and the number it partitions is <m>k+1+2+\cdots+n-1=k+\binom{n}{2}</m>. The proof that it is a bijection is the fact that subtracting <m>n-i</m> from the <m>i</m>\/th part of a partition of <m>k</m> into distinct parts yields a partition of <m>k</m>, because part <m>i+j</m> is at least <m>j</m> smaller than part <m>i</m>.
+          The number of partitions of <m>k</m> into <m>n</m> parts is equal to the number of partitions of <m>k+\binom{n}{2}</m> into <m>n</m> distinct parts.  The bijection from partitions of <m>k</m> with <m>n</m> parts to partitions of <m>k+\binom{n}{2}</m> with <m>n</m> distinct parts that proves this is the one that takes a partition <m>\lambda_n\lambda_{n-1}\cdots\lambda_1</m> of <m>k</m> with <m>\lambda_i>\lambda_{i+1}</m> and adds <m>i-1</m> to <m>\lambda_i</m> to get <m>\lambda'_i</m>. Then <m>\lambda'</m> is a partition into distinct parts, and the number it partitions is <m>k+1+2+\cdots+n-1=k+\binom{n}{2}</m>. The proof that it is a bijection is the fact that subtracting <m>n-i</m> from the <m>i</m>\/th part of a partition of <m>k</m> into distinct parts yields a partition of <m>k</m>, because part <m>i+j</m> is at least <m>j</m> smaller than part <m>i</m>.
         </p>
       </solution>
     </activity>

mbx/s4-2-genfn-int-parts.mbx

@@ -360,7 +360,7 @@ of it as <m>(2i) \cdot j</m> or if you think of it as <m>i \cdot (2j)</m>.
             </task>
             <task>
                 <statement>
-                    <p>How are <m>\qchoose{m+n}{n}</m> and <m>\qchoose{m+n}{n}</m> related? Prove it. (Note this is the same as asking how <m>\qchoose{r}{s}</m> and <m>\qchoose{r}{r-s}</m> are related.)
+                    <p>How are <m>\qchoose{m+n}{n}</m> and <m>\qchoose{m+n}{m}</m> related? Prove it. (Note this is the same as asking how <m>\qchoose{r}{s}</m> and <m>\qchoose{r}{r-s}</m> are related.)
                     </p>
                 </statement>
                 <hint>

mbx/s4-3-genfn-recurrence.mbx

@@ -30,7 +30,7 @@
             </me>
             and in the right hand side, use the fact that
             <me>
-              \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
             </me>
             (where we substituted
@@ -147,7 +147,7 @@
 
     <subsection>
         <title>Fibonacci numbers</title>
-        <p>The sequence of problems that follows (culminating in <xref ref="solveFibonacci" />) 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 Fibonacci<fn>Apparently Leanardo de Pisa was given the name Fibonacci posthumously. It is a shortening of <q>son of Bonacci</q> in Italian.</fn> introduced them, he did so with a fictional population of rabbits.<idx><h>Fibonacci numbers</h></idx>
+        <p>The sequence of problems that follows (culminating in <xref ref="solveFibonacci" />) 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 Fibonacci<fn>Apparently Leonardo da Pisa was given the name Fibonacci posthumously. It is a shortening of <q>son of Bonacci</q> in Italian.</fn> introduced them, he did so with a fictional population of rabbits.<idx><h>Fibonacci numbers</h></idx>
         </p>
     </subsection>
   <subsection>

mbx/s5-1-sizeunion.mbx

@@ -492,7 +492,7 @@
       <statement>
         <p>
           Frequently when we apply the principle of inclusion and exclusion, we will have a situation like that of part (d) of <xref ref="hatcheckprobpart">Problem</xref>.  That is, we will have a set <m>A</m> and subsets <m>A_1, A_2, \ldots, A_n</m> and we will want the size or the probability of the set of elements in <m>A</m> that are <em>not</em> in the union.  This set is known as the <term>complement</term> <idx><h>complement</h></idx> of the union of the <m>A_i</m>s in <m>A</m>, and is denoted by <m>A \setminus \bigcup_{i=1}^n A_i</m>, or if <m>A</m> is clear from context, by <m>\overline{\bigcup_{i=1}^n A_i}</m>.
-          Give the fomula for <m>\overline{\bigcup_{i=1}^n A_i}</m>.  The principle of inclusion and exclusion generall refers to both this formula and the one for the union.
+          Give the fomula for <m>\overline{\bigcup_{i=1}^n A_i}</m>.  The principle of inclusion and exclusion generally refers to both this formula and the one for the union.
         </p>
       </statement>
       <solution>