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 ## Student Responses
 
-<!-- Write your solutions below. Add sections for each problem as needed. -->
-<!-- Use proper mathematical notation where applicable. -->
+---
 
-[Your solutions here]
+## Part 1
 
+### Problem 1
+
+**How many bit strings of length 10 either start with 000 or end with 111?**
+
+Let A = bit strings of length 10 starting with 000.
+Let B = bit strings of length 10 ending with 111.
+
+- |A| = 2^7 = 128 (first 3 bits fixed, remaining 7 are free)
+- |B| = 2^7 = 128 (last 3 bits fixed, remaining 7 are free)
+- |A ∩ B| = 2^4 = 16 (first 3 and last 3 bits all fixed, middle 4 are free)
+
+By inclusion-exclusion:
+
+|A ∪ B| = |A| + |B| − |A ∩ B| = 128 + 128 − 16 = **240**
+
+---
+
+### Problem 2
+
+**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?**
+
+Represent each number as a 6-digit string (with leading zeros allowed), ranging from 000000 to 999999. Since we need digit sum 13, the value 000000 is automatically excluded (digit sum 0), so all counted solutions are positive integers.
+
+We need exactly one digit equal to 9 and total digit sum 13.
+
+- Choose which of the 6 positions holds the 9: C(6,1) = 6 ways.
+- The remaining 5 digits must each be in {0, 1, …, 8} and sum to 13 − 9 = 4.
+- Since the remaining digits sum to only 4, no digit can reach 9, so the constraint d_i ≤ 8 is automatically satisfied.
+- Number of non-negative integer solutions to d_1 + d_2 + d_3 + d_4 + d_5 = 4 with no upper bound constraint: C(4 + 5 − 1, 5 − 1) = C(8, 4) = 70.
+
+Total = 6 × 70 = **420**
+
+---
+
+### Problem 3
+
+**How many solutions are there to x_1 + x_2 + x_3 ≤ 11, where x_1, x_2, x_3 are nonneg integers?**
+
+Introduce a slack variable x_4 ≥ 0 to convert the inequality to an equality:
+
+x_1 + x_2 + x_3 + x_4 = 11
+
+The number of non-negative integer solutions is:
+
+C(11 + 4 − 1, 4 − 1) = C(14, 3) = 14 · 13 · 12 / 6 = **364**
+
+---
+
+### Problem 4
+
+**Prove, without using induction, that in any set of n + 1 positive integers not exceeding 2n there must be two that are relatively prime.**
+
+**Proof.** Partition the set {1, 2, …, 2n} into n pairs of consecutive integers:
+
+{1, 2}, {3, 4}, {5, 6}, …, {2n−1, 2n}.
+
+These n pairs are disjoint and cover all of {1, 2, …, 2n}.
+
+Given any set S of n + 1 positive integers not exceeding 2n, by the Pigeonhole Principle, at least two elements of S must belong to the same pair {2k−1, 2k} for some k.
+
+Call these two elements 2k−1 and 2k. They are consecutive integers, so gcd(2k−1, 2k) = 1: if d divides both 2k−1 and 2k, then d divides (2k) − (2k−1) = 1, so d = 1.
+
+Therefore, the set S contains two elements that are relatively prime. □
+
+---
+
+## Part 2
+
+### Problem 1
+
+**What is the number of positive integers not exceeding n = pq that are relatively prime to n?**
+
+This is Euler's totient function φ(pq). We count by exclusion.
+
+Among {1, 2, …, pq}, integers **not** coprime to pq are those divisible by p or by q.
+
+- Multiples of p in {1, …, pq}: p, 2p, …, qp → q multiples
+- Multiples of q in {1, …, pq}: q, 2q, …, pq → p multiples
+- Multiples of both p and q (i.e., multiples of pq): just pq → 1 multiple
+
+By inclusion-exclusion, integers not coprime to pq: q + p − 1.
+
+Therefore:
+
+φ(pq) = pq − (p + q − 1) = pq − p − q + 1 = **(p − 1)(q − 1)**
+
+---
+
+### Problem 2
+
+Assume n students and m instructors, with one designated lead instructor L. Total people: n + m.
+
+#### Part (a)
+
+**All students standing next to each other (no instructors between them).**
+
+Treat all n students as a single block. We then have m instructors + 1 student block = m + 1 objects to arrange in a line.
+
+- Arrange m + 1 objects: (m + 1)! ways
+- Arrange n students within the block: n! ways
+
+Total: **(m + 1)! · n!**
+
+---
+
+#### Part (b)
+
+**No two instructors standing next to each other.**
+
+First arrange the n students in a line: n! ways. This creates n + 1 gaps (including the two ends):
+
+_ S _ S _ … _ S _
+
+To ensure no two instructors are adjacent, place at most one instructor per gap. Choose m of the n + 1 gaps and assign the m distinct instructors (in order):
+
+- Choose m gaps from n + 1: C(n + 1, m) ways
+- Arrange m instructors in the chosen gaps: m! ways
+
+Total: **n! · C(n + 1, m) · m!**
+
+---
+
+#### Part (c)
+
+**No student is standing next to the lead instructor L.**
+
+Use inclusion-exclusion. Let A_i = set of all arrangements where student i is adjacent to L (for i = 1, …, n). We want (n+m)! − |A_1 ∪ … ∪ A_n|.
+
+**Computing |A_i|:** Treat {student i, L} as a block with 2 internal orderings. Then we have n + m − 1 units. |A_i| = 2 · (n + m − 1)!.
+
+**Computing |A_i ∩ A_j| for i ≠ j:** Both students i and j must be adjacent to L, meaning L is between them. The block (i–L–j or j–L–i) has 2 orderings and gives n + m − 2 units. |A_i ∩ A_j| = 2 · (n + m − 2)!.
+
+**Computing |A_i ∩ A_j ∩ A_k|:** L can have at most 2 neighbors, so three students cannot all be adjacent to L simultaneously. This is 0.
+
+By inclusion-exclusion:
+
+|A_1 ∪ … ∪ A_n| = C(n,1) · 2(n+m−1)! − C(n,2) · 2(n+m−2)!
+               = 2n(n+m−1)! − n(n−1)(n+m−2)!
+
+Valid arrangements = (n+m)! − 2n(n+m−1)! + n(n−1)(n+m−2)!
+
+Factor out (n+m−2)!:
+
+= (n+m−2)! · [(n+m)(n+m−1) − 2n(n+m−1) + n(n−1)]
+
+Expand the bracket:
+- (n+m)(n+m−1) = n² + 2mn + m² − n − m
+- 2n(n+m−1) = 2n² + 2mn − 2n
+- n(n−1) = n² − n
+
+Sum: (n² + 2mn + m² − n − m) − (2n² + 2mn − 2n) + (n² − n) = m² − m = m(m−1)
+
+Total: **m(m − 1)(n + m − 2)!**
+
+---
+
+## Part 3
+
+### Problem 1
+
+**Place 80 identical balls in 5 distinct bins, no bin has more than 24 balls.**
+
+Let x_i = number of balls in bin i. We need x_1 + x_2 + x_3 + x_4 + x_5 = 80 with 0 ≤ x_i ≤ 24.
+
+Without restriction: C(84, 4) ways.
+
+Let A_i = event that bin i has ≥ 25 balls. Use inclusion-exclusion to subtract violations.
+
+**|A_i|:** Substitute y_i = x_i − 25 ≥ 0. Sum becomes 55. Count = C(59, 4).
+There are C(5,1) = 5 such terms.
+
+**|A_i ∩ A_j|:** Two bins each ≥ 25, sum becomes 30. Count = C(34, 4).
+There are C(5,2) = 10 such terms.
+
+**|A_i ∩ A_j ∩ A_k|:** Three bins each ≥ 25, sum becomes 5. Count = C(9, 4).
+There are C(5,3) = 10 such terms.
+
+**|A_i ∩ A_j ∩ A_k ∩ A_l|:** Four bins each ≥ 25 requires at least 100 balls > 80. Impossible. Count = 0.
+
+Computing the binomial coefficients:
+
+- C(84, 4) = (84 · 83 · 82 · 81) / 24 = 46,308,024 / 24 = 1,929,501
+- C(59, 4) = (59 · 58 · 57 · 56) / 24 = 10,923,024 / 24 = 455,126
+- C(34, 4) = (34 · 33 · 32 · 31) / 24 = 1,113,024 / 24 = 46,376
+- C(9, 4) = (9 · 8 · 7 · 6) / 24 = 3024 / 24 = 126
+
+By inclusion-exclusion:
+
+|valid| = 1,929,501 − 5(455,126) + 10(46,376) − 10(126)
+        = 1,929,501 − 2,275,630 + 463,760 − 1,260
+        = **116,371**
+
+---
+
+### Problem 2
+
+**Permutations of INDISTINGUISHABLE such that GIT, TIN, and NAB do not appear.**
+
+**Step 1: Count letters in INDISTINGUISHABLE.**
+
+I–N–D–I–S–T–I–N–G–U–I–S–H–A–B–L–E (17 letters total):
+- I × 4, N × 2, S × 2, D × 1, T × 1, G × 1, U × 1, H × 1, A × 1, B × 1, L × 1, E × 1
+
+Total permutations = 17! / (4! · 2! · 2!)
+
+**Step 2: Define events and overlap analysis.**
+
+Let A = permutations containing "GIT", B = permutations containing "TIN", C = permutations containing "NAB".
+
+Key observations about overlaps (using letter multiplicities):
+- **T appears only once.** Both GIT and TIN require T, so if both appear, they must share T. The only overlap is GIT + TIN sharing T → **GITIN** (G–I–T–I–N).
+- **TIN ends with N; NAB starts with N.** They can share N → **TINAB** (T–I–N–A–B), or use separate N's (N appears twice).
+- **GIT and NAB share no letters**, so they always appear as separate blocks.
+
+**Step 3: Compute each term.**
+
+When treating a substring as a block, remove its letters from the pool and count permutations of (remaining letters + block) with appropriate repetition factors.
+
+**|A|:** GIT block. Remaining: I×3, N×2, S×2, others×1. Units = 15.
+|A| = 15! / (3! · 2! · 2!) = 54,486,432,000
+
+**|B|:** TIN block. Remaining: I×3, N×1, S×2, others×1. Units = 15.
+|B| = 15! / (3! · 2!) = 108,972,864,000
+
+**|C|:** NAB block. Remaining: I×4, N×1, S×2, others×1. Units = 15.
+|C| = 15! / (4! · 2!) = 27,243,216,000
+
+**|A ∩ B|:** GIT and TIN must overlap as GITIN (only one T). GITIN block uses G×1, I×2, T×1, N×1. Remaining: I×2, N×1, S×2, others×1. Units = 13.
+|A ∩ B| = 13! / (2! · 2!) = 1,556,755,200
+
+**|A ∩ C|:** GIT block + NAB block (no shared letters). Remaining: I×3, N×1, S×2, others×1. Units = 13.
+|A ∩ C| = 13! / (3! · 2!) = 518,918,400
+
+**|B ∩ C|:** Two sub-cases (both valid since N appears twice):
+- *Case 1 (overlap):* TINAB block (sharing N). Remaining: I×3, N×1, S×2, G×1, others×1. Units = 13. Count = 13! / (3! · 2!) = 518,918,400.
+- *Case 2 (non-overlap):* TIN block + NAB block (each using its own N). Remaining: I×3, S×2, G×1, others×1. Units = 13. Count = 13! / (3! · 2!) = 518,918,400.
+
+|B ∩ C| = 2 × 518,918,400 = 1,037,836,800
+
+**|A ∩ B ∩ C|:** GIT and TIN must form GITIN; then NAB can share or not share N with GITIN:
+- *Case 1 (overlap):* GITINAB block (G–I–T–I–N–A–B). Remaining: I×2, N×1, S×2, others×1. Units = 11. Count = 11! / (2! · 2!) = 9,979,200.
+- *Case 2 (non-overlap):* GITIN block + NAB block (using the remaining N). Remaining: I×2, S×2, others×1. Units = 11. Count = 11! / (2! · 2!) = 9,979,200.
+
+|A ∩ B ∩ C| = 2 × 9,979,200 = 19,958,400
+
+**Step 4: Apply inclusion-exclusion.**
+
+|A ∪ B ∪ C| = |A| + |B| + |C| − |A∩B| − |A∩C| − |B∩C| + |A∩B∩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
+
+Total permutations = 17! / (4! · 2! · 2!) = 3,705,077,376,000
+
+Answer = 3,705,077,376,000 − 187,608,960,000 = **3,517,468,416,000**
+
+---
+
+### Problem 3
+
+**Number of 1-1 functions f : {1, …, 20} → {1, …, 80} with no fixed point.**
+
+A fixed point is a ∈ {1, …, 20} with f(a) = a (only elements 1 through 20 can be fixed points, since a must lie in both the domain and the codomain for f(a) = a to hold, and the domain is {1,…,20}).
+
+**Total injective functions:** P(80, 20) = 80! / 60!
+
+**Apply inclusion-exclusion.** Let A_i = set of injective functions where f(i) = i.
+
+**|A_i|:** Fix f(i) = i; map remaining 19 domain elements injectively into 79 remaining codomain elements.
+|A_i| = P(79, 19) = 79! / 60!
+
+**|A_i ∩ A_j| (i ≠ j):** Fix f(i) = i and f(j) = j; remaining 18 elements map into 78 elements.
+|A_i ∩ A_j| = P(78, 18) = 78! / 60!
+
+**General k-fold intersection:** Fix f(i_1) = i_1, …, f(i_k) = i_k; remaining 20−k elements map into 80−k elements.
+|A_{i_1} ∩ … ∩ A_{i_k}| = P(80−k, 20−k) = (80−k)! / 60!
+
+By inclusion-exclusion, the number of injective functions with **at least one** fixed point is:
+
+sum_{k=1}^{20} (−1)^{k+1} · C(20, k) · (80−k)! / 60!
+
+Therefore, the number of injective functions with **no fixed point** is:
+
+sum_{k=0}^{20} (−1)^k · C(20, k) · P(80−k, 20−k)
+
+= sum_{k=0}^{20} (−1)^k · C(20, k) · (80−k)! / 60!
+
+Expanding:
+
+= P(80,20) − C(20,1)·P(79,19) + C(20,2)·P(78,18) − … + (−1)^{20}·C(20,20)·P(60,0)
+
+= (1/60!) · sum_{k=0}^{20} (−1)^k · C(20, k) · (80−k)!