1. **Problem 1a:** Given today is Sunday and neither this year nor next is a leap year, find the day of the week one year from today. - Since a non-leap year has 365 days, and 365 mod 7 = 1, the day advances by 1 weekday. - Today is Sunday, so one year from today will be Monday. 2. **Problem 1b:** Compute: i. $32 \div 9$ - Division gives quotient $\lfloor \frac{32}{9} \rfloor = 3$. ii. $32 \bmod 9$ - Remainder is $32 - 9 \times 3 = 32 - 27 = 5$. 3. **Problem 1c:** Given integer $m$ with $m \bmod 11 = 6$, compute $4m \bmod 11$. - Since $m \equiv 6 \pmod{11}$, multiply both sides by 4: $$4m \equiv 4 \times 6 = 24 \pmod{11}$$ - Simplify $24 \bmod 11$: $$24 \bmod 11 = 24 - 2 \times 11 = 24 - 22 = 2$$ - Therefore, $4m \bmod 11 = 2$. 4. **Problem 2a:** Show that $\binom{n}{n-2} = \frac{n(n-1)}{2}$ for $n \geq 2$. - Recall the binomial coefficient formula: $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$ - Substitute $k = n-2$: $$\binom{n}{n-2} = \frac{n!}{(n-2)!2!}$$ - Simplify numerator: $$n! = n \times (n-1) \times (n-2)!$$ - Substitute back: $$\binom{n}{n-2} = \frac{n \times (n-1) \times (n-2)!}{(n-2)! \times 2} = \frac{n(n-1)}{2}$$ 5. **Problem 3a:** From 12 students (5 men, 7 women), find number of 5-person teams with at least one man. - Total 5-person teams: $$\binom{12}{5} = \frac{12!}{5!7!} = 792$$ - Teams with no men (all women): $$\binom{7}{5} = \frac{7!}{5!2!} = 21$$ - Teams with at least one man: $$792 - 21 = 771$$ 6. **Problem 3b:** Survey of 270 students: - Likes meat (M): 64 - Likes eggs (E): 94 - Likes vegetables (V): 58 - Likes M and E: 26 - Likes M and V: 28 - Likes E and V: 22 - Likes all three: 14 - Use inclusion-exclusion principle to find number who like at least one: $$|M \cup E \cup V| = |M| + |E| + |V| - |M \cap E| - |M \cap V| - |E \cap V| + |M \cap E \cap V|$$ $$= 64 + 94 + 58 - 26 - 28 - 22 + 14 = 216$$ - Number who do not like any: $$270 - 216 = 54$$ 7. **Problem 3c:** Find number of integers $\leq 1000$ that are neither multiples of 3 nor 5. - Count multiples of 3: $$\left\lfloor \frac{1000}{3} \right\rfloor = 333$$ - Count multiples of 5: $$\left\lfloor \frac{1000}{5} \right\rfloor = 200$$ - Count multiples of 15 (LCM of 3 and 5): $$\left\lfloor \frac{1000}{15} \right\rfloor = 66$$ - Use inclusion-exclusion for multiples of 3 or 5: $$333 + 200 - 66 = 467$$ - Numbers not multiples of 3 or 5: $$1000 - 467 = 533$$ 8. **Problem 4a:** Prove for nonnegative integer $n$: $$\sum_{k=0}^n \binom{n}{k} 3^k = 4^n$$ - Use binomial theorem: $$(1 + x)^n = \sum_{k=0}^n \binom{n}{k} x^k$$ - Substitute $x=3$: $$\sum_{k=0}^n \binom{n}{k} 3^k = (1 + 3)^n = 4^n$$ 9. **Problem 4b:** Find coefficient of $x^9 y^{10}$ in expansion of $(2x - 3y^2)^{14}$. - General term in binomial expansion: $$\binom{14}{k} (2x)^k (-3y^2)^{14-k} = \binom{14}{k} 2^k (-3)^{14-k} x^k y^{2(14-k)}$$ - Powers of $x$ and $y$: $$x^k y^{28 - 2k}$$ - Set powers equal to desired powers: $$k = 9$$ $$28 - 2k = 10 \implies 28 - 2 \times 9 = 10$$ - Both conditions satisfied at $k=9$. - Coefficient: $$\binom{14}{9} 2^9 (-3)^5 = \binom{14}{9} 2^9 (-243)$$ - Calculate $\binom{14}{9} = \binom{14}{5} = \frac{14 \times 13 \times 12 \times 11 \times 10}{5 \times 4 \times 3 \times 2 \times 1} = 2002$ - Calculate $2^9 = 512$ - Multiply: $$2002 \times 512 \times (-243) = -249,034,752$$ - Final coefficient is $-249034752$. **Final answers:** 1a. Monday 1b.i. 3 1b.ii. 5 1c. 2 2a. $\frac{n(n-1)}{2}$ 3a. 771 3b. 54 3c. 533 4a. $4^n$ 4b. $-249034752$