1. **QUESTION ONE** 1.a. Problem: 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 (because 7*52=364, remainder 1), the day advances by 1 weekday. - Therefore, one year from Sunday is Monday. 1.b. Compute: i. $32 \div 9$ (integer division): - $9 \times 3 = 27$ and $9 \times 4 = 36 > 32$, so quotient is 3. ii. $32 \bmod 9$ (remainder): - $32 - 9 \times 3 = 32 - 27 = 5$. 1.c. Given $m \bmod 11 = 6$, compute $4m \bmod 11$. - Since $m \equiv 6 \pmod{11}$, multiply both sides by 4: $$4m \equiv 4 \times 6 = 24 \pmod{11}$$ - Simplify $24 \bmod 11$: $$24 - 2 \times 11 = 24 - 22 = 2$$ - So, $4m \bmod 11 = 2$. 2. **QUESTION TWO** 2.a. 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: $$\binom{n}{n-2} = \frac{n \times (n-1) \times (n-2)!}{(n-2)! \times 2} = \frac{n(n-1)}{2}$$ 3. **QUESTION THREE** 3.a. Number of 5-person teams from 12 students (5 men, 7 women) with at least one man. - Total 5-person teams: $$\binom{12}{5}$$ - Teams with no men (all women): $$\binom{7}{5}$$ - Teams with at least one man: $$\binom{12}{5} - \binom{7}{5}$$ - Calculate: $$\binom{12}{5} = \frac{12!}{5!7!} = 792$$ $$\binom{7}{5} = \frac{7!}{5!2!} = 21$$ - Result: $$792 - 21 = 771$$ 3.b. Survey of 270 students: 64 like meat (M), 94 like eggs (E), 58 like vegetables (V), 26 like M and E, 28 like M and V, 22 like E and V, 14 like all three. - Use Inclusion-Exclusion Principle: $$|M \cup E \cup V| = |M| + |E| + |V| - |M \cap E| - |M \cap V| - |E \cap V| + |M \cap E \cap V|$$ - Substitute values: $$64 + 94 + 58 - 26 - 28 - 22 + 14 = 216 - 76 + 14 = 154$$ - Number who do not like any: $$270 - 154 = 116$$ 3.c. Number of integers $\leq 1000$ that are neither multiples of 3 nor 5. - Total integers: 1000 - Multiples of 3: $$\left\lfloor \frac{1000}{3} \right\rfloor = 333$$ - Multiples of 5: $$\left\lfloor \frac{1000}{5} \right\rfloor = 200$$ - Multiples of 15 (both 3 and 5): $$\left\lfloor \frac{1000}{15} \right\rfloor = 66$$ - Using Inclusion-Exclusion: $$\text{Multiples of 3 or 5} = 333 + 200 - 66 = 467$$ - Numbers neither multiples of 3 nor 5: $$1000 - 467 = 533$$ 4. **QUESTION FOUR** 4.a. 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 $x=3$: $$(1 + 3)^n = \sum_{k=0}^n \binom{n}{k} 3^k$$ - Simplify left side: $$4^n = \sum_{k=0}^n \binom{n}{k} 3^k$$ 4.b. Find coefficient of $x^9 y^{10}$ in expansion of $(2x - 3y^2)^{14}$. - General term of binomial expansion: $$T_{k+1} = \binom{14}{k} (2x)^{14-k} (-3y^2)^k$$ - Powers: $$x^{14-k} y^{2k}$$ - We want: $$14 - k = 9 \Rightarrow k = 5$$ $$2k = 10$$ - Coefficient: $$\binom{14}{5} (2)^{9} (-3)^{5}$$ - Calculate: $$\binom{14}{5} = 2002$$ $$2^{9} = 512$$ $$(-3)^5 = -243$$ - Multiply: $$2002 \times 512 \times (-243) = -249,125,632$$ **Final answers:** 1.a. Monday 1.b.i. 3 1.b.ii. 5 1.c. 2 2.a. $\frac{n(n-1)}{2}$ 3.a. 771 3.b. 116 3.c. 533 4.a. $4^n$ 4.b. $-249125632$