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-# COMP182 Homework 7
+# COMP182 Homework 7: Counting
 
-## Student Responses
+## Part 1
 
-<!-- Write your solutions below. Add sections for each problem as needed. -->
-<!-- Use proper mathematical notation where applicable. -->
+### Problem 1 [7 pts]
 
-[Your solutions here]
+**How many bit strings of length 10 either start with 000 or end with 111?**
 
+We use the inclusion-exclusion principle.
+
+Let $A$ = set of bit strings of length 10 that start with 000.
+Let $B$ = set of bit strings of length 10 that end with 111.
+
+- $|A|$: The first 3 bits are fixed as 000, and the remaining 7 bits can be anything. So $|A| = 2^7 = 128$.
+- $|B|$: The last 3 bits are fixed as 111, and the remaining 7 bits can be anything. So $|B| = 2^7 = 128$.
+- $|A \cap B|$: The first 3 bits are 000 and the last 3 bits are 111. The middle 4 bits can be anything. So $|A \cap B| = 2^4 = 16$.
+
+By inclusion-exclusion:
+
+$$|A \cup B| = |A| + |B| - |A \cap B| = 128 + 128 - 16 = \boxed{240}$$
+
+---
+
+### Problem 2 [7 pts]
+
+**How many positive integers less than 1,000,000 have exactly one digit equal to 9 and have a sum of digits equal to 13?**
+
+Positive integers less than 1,000,000 have at most 6 digits. We represent each such integer as a 6-digit string $(d_1, d_2, d_3, d_4, d_5, d_6)$ with leading zeros allowed, where $0 \le d_i \le 9$ and $d_1 + d_2 + \cdots + d_6 = 13$. (Note: 000000 = 0 is excluded, but it does not satisfy our conditions anyway.)
+
+**Step 1: Choose which digit equals 9.**
+There are $\binom{6}{1} = 6$ ways to choose which one of the 6 positions has the digit 9.
+
+**Step 2: Count arrangements of the remaining 5 digits.**
+The remaining 5 digits must satisfy:
+- Sum to $13 - 9 = 4$
+- Each digit is between 0 and 8 (not 9, since exactly one digit equals 9)
+
+Since the sum is 4 and each digit is at most 8, the upper bound constraint $d_i \le 8$ is automatically satisfied (the maximum any single digit can be is 4). So we just need the number of nonneg integer solutions to $e_1 + e_2 + e_3 + e_4 + e_5 = 4$.
+
+By stars and bars: $\binom{4 + 5 - 1}{5 - 1} = \binom{8}{4} = 70$.
+
+**Total:** $6 \times 70 = \boxed{420}$
+
+---
+
+### Problem 3 [7 pts]
+
+**How many solutions are there to $x_1 + x_2 + x_3 \le 11$, where $x_1, x_2, x_3$ are nonneg integers?**
+
+Introduce a slack variable $x_4 \ge 0$ to convert the inequality to an equality:
+
+$$x_1 + x_2 + x_3 + x_4 = 11$$
+
+where $x_1, x_2, x_3, x_4$ are all nonneg integers.
+
+By stars and bars, the number of solutions is:
+
+$$\binom{11 + 4 - 1}{4 - 1} = \binom{14}{3} = \frac{14 \times 13 \times 12}{3!} = \frac{2184}{6} = \boxed{364}$$
+
+---
+
+### Problem 4 [8 pts]
+
+**Prove that in any set of $n+1$ positive integers not exceeding $2n$, there must be two that are relatively prime.**
+
+**Proof.** Partition the integers $\{1, 2, \ldots, 2n\}$ into $n$ pairs of consecutive integers:
+
+$$\{1, 2\}, \{3, 4\}, \{5, 6\}, \ldots, \{2n-1, 2n\}.$$
+
+There are exactly $n$ such pairs. If we choose any $n+1$ integers from $\{1, 2, \ldots, 2n\}$, then by the **Pigeonhole Principle**, at least two of the chosen integers must belong to the same pair $\{2k-1, 2k\}$ for some $k$.
+
+These two integers are consecutive, namely $2k-1$ and $2k$. We claim they are relatively prime. Indeed, if $d = \gcd(2k-1, 2k)$, then $d$ divides both $2k$ and $2k-1$, so $d$ divides their difference:
+
+$$d \mid (2k - (2k-1)) = 1.$$
+
+Therefore $d = 1$, and the two integers are relatively prime.
+
+Since any set of $n+1$ positive integers not exceeding $2n$ must contain two consecutive integers, it must contain two integers that are relatively prime. $\blacksquare$
+
+---
+
+## Part 2
+
+### Problem 1 [8 pts]
+
+**Given distinct primes $p$ and $q$ with $n = pq$, find the number of positive integers not exceeding $n$ that are relatively prime to $n$.**
+
+This is Euler's totient function $\varphi(n) = \varphi(pq)$.
+
+A positive integer $k \le n = pq$ is **not** relatively prime to $n$ if and only if $p \mid k$ or $q \mid k$.
+
+By inclusion-exclusion, the number of integers in $\{1, \ldots, pq\}$ divisible by $p$ or $q$ is:
+
+- Multiples of $p$ in $\{1, \ldots, pq\}$: there are $\frac{pq}{p} = q$ of them.
+- Multiples of $q$ in $\{1, \ldots, pq\}$: there are $\frac{pq}{q} = p$ of them.
+- Multiples of $pq$ in $\{1, \ldots, pq\}$: there is $\frac{pq}{pq} = 1$ of them.
+
+Number divisible by $p$ or $q$ $= q + p - 1$.
+
+Therefore:
+
+$$\varphi(pq) = pq - (p + q - 1) = pq - p - q + 1 = \boxed{(p-1)(q-1)}$$
+
+---
+
+### Problem 2(a) [7 pts]
+
+**How many ways to arrange $n$ students and $m$ instructors in a line such that all students are standing next to each other?**
+
+Treat the $n$ students as a single block. Then we have $(m + 1)$ entities to arrange in a line: the student block and $m$ instructors.
+
+- These $(m+1)$ entities can be arranged in $(m+1)!$ ways.
+- Within the student block, the $n$ students can be arranged in $n!$ ways.
+
+$$\text{Total} = \boxed{(m+1)! \cdot n!}$$
+
+---
+
+### Problem 2(b) [7 pts]
+
+**How many ways to arrange $n$ students and $m$ instructors in a line such that no two instructors are standing next to each other?**
+
+**Step 1:** Arrange the $n$ students in a line: $n!$ ways.
+
+**Step 2:** This creates $(n+1)$ gaps (including the two ends):
+
+$$\_ \; S_1 \; \_ \; S_2 \; \_ \; \cdots \; \_ \; S_n \; \_$$
+
+**Step 3:** Choose $m$ of these $(n+1)$ gaps in which to place the $m$ instructors (at most one per gap ensures no two instructors are adjacent): $\binom{n+1}{m}$ ways.
+
+**Step 4:** Arrange the $m$ instructors in the chosen gaps: $m!$ ways.
+
+$$\text{Total} = n! \cdot \binom{n+1}{m} \cdot m! = \boxed{\frac{n! \cdot (n+1)!}{(n+1-m)!}}$$
+
+---
+
+### Problem 2(c) [7 pts]
+
+**How many ways to arrange $n$ students and $m$ instructors in a line such that no student is standing next to the lead instructor?**
+
+Let $L$ denote the lead instructor. We count directly by considering where $L$ is placed and ensuring both neighbors of $L$ (if they exist) are instructors, not students.
+
+**Case 1: $L$ is at one of the two ends (position 1 or position $n+m$).**
+$L$ has exactly one neighbor, which must be one of the $(m-1)$ other instructors.
+- 2 choices for which end $L$ occupies.
+- $(m-1)$ choices for which instructor is $L$'s neighbor.
+- The remaining $(n+m-2)$ people can be arranged in the remaining positions in $(n+m-2)!$ ways.
+- Count: $2(m-1)(n+m-2)!$
+
+**Case 2: $L$ is at an interior position.**
+$L$ has two neighbors, both of which must be instructors.
+- $(n+m-2)$ choices for $L$'s interior position.
+- Choose 2 of the $(m-1)$ other instructors for $L$'s neighbors, with order (left vs. right): $P(m-1, 2) = (m-1)(m-2)$ ways.
+- The remaining $(n+m-3)$ people fill the remaining positions: $(n+m-3)!$ ways.
+- Count: $(n+m-2)(m-1)(m-2)(n+m-3)! = (m-1)(m-2)(n+m-2)!$
+
+**Total:**
+
+$$2(m-1)(n+m-2)! + (m-1)(m-2)(n+m-2)!$$
+$$= (m-1)(n+m-2)!\bigl[2 + (m-2)\bigr]$$
+$$= \boxed{m(m-1)(n+m-2)!}$$
+
+*Verification with $n=1, m=2$: The formula gives $2 \cdot 1 \cdot 1! = 2$. The valid arrangements of (S, L, I) with no student next to L are: S-I-L and L-I-S, which is indeed 2.* $\checkmark$
+
+---
+
+## Part 3
+
+### Problem 1 [14 pts]
+
+**How many ways to place 80 identical balls in 5 distinct bins such that no bin has more than 24 balls?**
+
+We need the number of nonneg integer solutions to $x_1 + x_2 + x_3 + x_4 + x_5 = 80$ with $0 \le x_i \le 24$ for all $i$.
+
+**Using inclusion-exclusion on the upper bound constraints:**
+
+Let $A_i$ be the set of solutions where $x_i \ge 25$. We want to subtract these "bad" solutions.
+
+For $|A_i|$: substitute $y_i = x_i - 25 \ge 0$, giving a sum of $80 - 25 = 55$ with 5 nonneg variables:
+$$|A_i| = \binom{55 + 4}{4} = \binom{59}{4}$$
+
+For $|A_i \cap A_j|$: two variables $\ge 25$, so sum becomes $80 - 50 = 30$:
+$$|A_i \cap A_j| = \binom{30 + 4}{4} = \binom{34}{4}$$
+
+For $|A_i \cap A_j \cap A_k|$: three variables $\ge 25$, sum becomes $80 - 75 = 5$:
+$$|A_i \cap A_j \cap A_k| = \binom{5 + 4}{4} = \binom{9}{4}$$
+
+For four or more: $80 - 100 < 0$, so impossible. These terms are 0.
+
+By inclusion-exclusion:
+
+$$\text{Answer} = \binom{84}{4} - \binom{5}{1}\binom{59}{4} + \binom{5}{2}\binom{34}{4} - \binom{5}{3}\binom{9}{4}$$
+
+Computing each term:
+
+- $\binom{84}{4} = \frac{84 \cdot 83 \cdot 82 \cdot 81}{24} = \frac{46{,}308{,}024}{24} = 1{,}929{,}501$
+
+- $\binom{59}{4} = \frac{59 \cdot 58 \cdot 57 \cdot 56}{24} = \frac{10{,}923{,}024}{24} = 455{,}126$
+
+- $\binom{34}{4} = \frac{34 \cdot 33 \cdot 32 \cdot 31}{24} = \frac{1{,}113{,}024}{24} = 46{,}376$
+
+- $\binom{9}{4} = 126$
+
+$$= 1{,}929{,}501 - 5(455{,}126) + 10(46{,}376) - 10(126)$$
+$$= 1{,}929{,}501 - 2{,}275{,}630 + 463{,}760 - 1{,}260$$
+$$= \boxed{116{,}371}$$
+
+---
+
+### Problem 2 [14 pts]
+
+**Number of permutations of the letters in INDISTINGUISHABLE such that the strings GIT, TIN, and NAB do not appear as consecutive substrings.**
+
+**Letter frequencies in INDISTINGUISHABLE (17 letters):**
+
+| Letter | Count |
+|--------|-------|
+| I | 4 |
+| N | 2 |
+| S | 2 |
+| A, B, D, E, G, H, L, T, U | 1 each |
+
+Total unrestricted permutations: $\frac{17!}{4! \cdot 2! \cdot 2!} = \frac{17!}{96}$
+
+We use inclusion-exclusion. Let:
+- $A$ = permutations containing GIT as a consecutive substring
+- $B$ = permutations containing TIN as a consecutive substring
+- $C$ = permutations containing NAB as a consecutive substring
+
+**Key observation:** Since we have only 1 copy of G, T, and B, each pattern can appear **at most once** in any permutation.
+
+**Single sets** (treat each substring as a block, reducing 17 items to 15):
+
+- $|A|$: Block GIT uses G(1), I(1), T(1). Remaining: I(3), N(2), S(2), plus 8 unique letters + block = 15 items.
+$$|A| = \frac{15!}{3! \cdot 2! \cdot 2!} = \frac{15!}{24}$$
+
+- $|B|$: Block TIN uses T(1), I(1), N(1). Remaining: I(3), N(1), S(2), plus 8 unique + block = 15 items.
+$$|B| = \frac{15!}{3! \cdot 2!} = \frac{15!}{12}$$
+
+- $|C|$: Block NAB uses N(1), A(1), B(1). Remaining: I(4), N(1), S(2), plus 7 unique + block = 15 items.
+$$|C| = \frac{15!}{4! \cdot 2!} = \frac{15!}{48}$$
+
+**Pairwise intersections:**
+
+- $|A \cap B|$ (both GIT and TIN): GIT and TIN share the letter T (only 1 T available), so they **cannot** appear separately. They must overlap, forming **GITIN** (G-I-T-I-N, sharing the T). Block uses G(1), I(2), T(1), N(1). Remaining: I(2), N(1), S(2), plus 7 unique + block = 13 items.
+$$|A \cap B| = \frac{13!}{2! \cdot 2!} = \frac{13!}{4}$$
+
+- $|A \cap C|$ (both GIT and NAB): GIT and NAB share no letters, so they cannot overlap. Two separate blocks. Used: G(1), I(1), T(1), N(1), A(1), B(1). Remaining: I(3), N(1), S(2), plus 6 unique + 2 blocks = 13 items.
+$$|A \cap C| = \frac{13!}{3! \cdot 2!} = \frac{13!}{12}$$
+
+- $|B \cap C|$ (both TIN and NAB): TIN ends with N, NAB starts with N. Two sub-cases:
+
+  **Sub-case 1 (non-overlapping):** TIN and NAB appear separately. Uses T(1), I(1), N(2), A(1), B(1). We have N(2), so this works. Remaining: I(3), S(2), plus 6 unique + 2 blocks = 13 items.
+  $$\frac{13!}{3! \cdot 2!} = \frac{13!}{12}$$
+
+  **Sub-case 2 (overlapping as TINAB):** Block uses T(1), I(1), N(1), A(1), B(1). Remaining: I(3), N(1), S(2), plus 6 unique + block = 13 items.
+  $$\frac{13!}{3! \cdot 2!} = \frac{13!}{12}$$
+
+  $$|B \cap C| = \frac{13!}{12} + \frac{13!}{12} = \frac{13!}{6}$$
+
+**Triple intersection $|A \cap B \cap C|$:**
+
+From above, $A \cap B$ requires the GITIN block. We also need NAB. GITIN ends with N, NAB starts with N, so again two sub-cases:
+
+**Sub-case 1 (non-overlapping GITIN and NAB):** Uses N(2) total. Remaining: I(2), S(2), plus 5 unique + 2 blocks = 11 items.
+$$\frac{11!}{2! \cdot 2!} = \frac{11!}{4}$$
+
+**Sub-case 2 (overlapping as GITINAB):** Uses N(1). Remaining: I(2), N(1), S(2), plus 5 unique + block = 11 items.
+$$\frac{11!}{2! \cdot 2!} = \frac{11!}{4}$$
+
+$$|A \cap B \cap C| = \frac{11!}{4} + \frac{11!}{4} = \frac{11!}{2}$$
+
+**Applying inclusion-exclusion:**
+
+$$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|$$
+
+Computing numerically:
+
+| Term | Expression | Value |
+|------|-----------|-------|
+| $\|A\|$ | $15!/24$ | $54{,}486{,}432{,}000$ |
+| $\|B\|$ | $15!/12$ | $108{,}972{,}864{,}000$ |
+| $\|C\|$ | $15!/48$ | $27{,}243{,}216{,}000$ |
+| $\|A \cap B\|$ | $13!/4$ | $1{,}556{,}755{,}200$ |
+| $\|A \cap C\|$ | $13!/12$ | $518{,}918{,}400$ |
+| $\|B \cap C\|$ | $13!/6$ | $1{,}037{,}836{,}800$ |
+| $\|A \cap B \cap C\|$ | $11!/2$ | $19{,}958{,}400$ |
+
+$$|A \cup B \cup C| = (54{,}486{,}432{,}000 + 108{,}972{,}864{,}000 + 27{,}243{,}216{,}000)$$
+$$- (1{,}556{,}755{,}200 + 518{,}918{,}400 + 1{,}037{,}836{,}800) + 19{,}958{,}400$$
+$$= 190{,}702{,}512{,}000 - 3{,}113{,}510{,}400 + 19{,}958{,}400 = 187{,}608{,}960{,}000$$
+
+**Answer:**
+
+$$\frac{17!}{96} - |A \cup B \cup C| = 3{,}705{,}077{,}376{,}000 - 187{,}608{,}960{,}000 = \boxed{3{,}517{,}468{,}416{,}000}$$
+
+---
+
+### Problem 3 [14 pts]
+
+**How many injective (1-1) functions $f : \{1, \ldots, 20\} \to \{1, \ldots, 80\}$ have no fixed point?**
+
+A fixed point exists when $f(a) = a$ for some $a \in \{1, \ldots, 20\}$ (note: since $\{1,\ldots,20\} \subset \{1,\ldots,80\}$, fixed points are possible for all domain elements).
+
+**Using inclusion-exclusion.** Let $A_i$ be the set of injections where $f(i) = i$, for $i = 1, \ldots, 20$.
+
+For any subset $S \subseteq \{1,\ldots,20\}$ with $|S| = k$: fix $f(i) = i$ for all $i \in S$. The remaining $20 - k$ elements must map injectively to $\{1,\ldots,80\} \setminus S$, which has $80 - k$ elements. The number of such injections is:
+
+$$P(80-k, 20-k) = \frac{(80-k)!}{60!}$$
+
+(since $(80-k) - (20-k) = 60$).
+
+There are $\binom{20}{k}$ choices for $S$, so:
+
+$$\left|\bigcap_{i \in S} A_i\right| = \frac{(80-k)!}{60!}$$
+
+By inclusion-exclusion, the number of injections with **no** fixed point is:
+
+$$\sum_{k=0}^{20} (-1)^k \binom{20}{k} \frac{(80-k)!}{60!}$$
+
+Equivalently, factoring out:
+
+$$= \frac{1}{60!} \sum_{k=0}^{20} (-1)^k \binom{20}{k} (80-k)!$$
+
+$$= \boxed{\sum_{k=0}^{20} (-1)^k \binom{20}{k} \frac{(80-k)!}{60!}}$$
+
+*Verification with small example: $f: \{1\} \to \{1,2\}$ with no fixed point. Formula gives $\binom{1}{0}\frac{2!}{1!} - \binom{1}{1}\frac{1!}{1!} = 2 - 1 = 1$, which is correct ($f(1) = 2$ is the only option).* $\checkmark$