1. **Stating the problem:** We are given probabilities expressed using combinations and limits involving sequences and exponential functions. 2. **Understanding combinations:** The combination formula is given by: $$ C_n^k = \frac{n!}{k!(n-k)!} $$ where $n!$ is the factorial of $n$. 3. **Calculating $P(A)$:** Given: $$ P(A) = \frac{3C_3 \times 4C_2}{7C_5} $$ Calculate each term: $$ 3C_3 = \frac{3!}{3!0!} = 1 $$ $$ 4C_2 = \frac{4!}{2!2!} = 6 $$ $$ 7C_5 = \frac{7!}{5!2!} = 21 $$ So, $$ P(A) = \frac{1 \times 6}{21} = \frac{6}{21} $$ Simplify by canceling common factor 3: $$ P(A) = \frac{\cancel{6}^2}{\cancel{21}^7} = \frac{2}{7} $$ 4. **Calculating $P(B)$:** Given: $$ P(B) = \frac{3C_3 \times 5C_2}{8C_5} $$ Calculate each term: $$ 3C_3 = 1 $$ $$ 5C_2 = \frac{5!}{2!3!} = 10 $$ $$ 8C_5 = \frac{8!}{5!3!} = 56 $$ So, $$ P(B) = \frac{1 \times 10}{56} = \frac{10}{56} $$ Simplify by canceling common factor 2: $$ P(B) = \frac{\cancel{10}^5}{\cancel{56}^{28}} = \frac{5}{28} $$ 5. **Calculating $P(C)$:** Given: $$ P(C) = \frac{3C_3 \times 6C_2}{9C_5} $$ Calculate each term: $$ 3C_3 = 1 $$ $$ 6C_2 = \frac{6!}{2!4!} = 15 $$ $$ 9C_5 = \frac{9!}{5!4!} = 126 $$ So, $$ P(C) = \frac{1 \times 15}{126} = \frac{15}{126} $$ Simplify by canceling common factor 21: $$ P(C) = \frac{\cancel{15}^1}{\cancel{126}^6} = \frac{1}{6} $$ 6. **Evaluating limits:** - $$ \lim_{n \to +\infty} n^2 + n = +\infty $$ because polynomial grows without bound. - $$ \lim_{n \to +\infty} 3^n = +\infty $$ exponential growth. - $$ \lim_{n \to +\infty} \left(\frac{2}{5}\right)^n = 0 $$ since base $<1$. - $$ \lim_{n \to +\infty} \frac{n^2 + 4n}{n^2 + 5} = 1 $$ divide numerator and denominator by $n^2$: $$ \frac{1 + \frac{4}{n}}{1 + \frac{5}{n^2}} \to \frac{1+0}{1+0} = 1 $$ - $$ \lim_{n \to +\infty} \frac{1 + 4n}{3n + 6} = \frac{4}{3} $$ divide numerator and denominator by $n$: $$ \frac{\frac{1}{n} + 4}{3 + \frac{6}{n}} \to \frac{0 + 4}{3 + 0} = \frac{4}{3} $$ - $$ \lim_{n \to +\infty} 2n^3 = +\infty $$ polynomial growth. **Final answers:** $$ P(A) = \frac{2}{7}, \quad P(B) = \frac{5}{28}, \quad P(C) = \frac{1}{6} $$ $$ \lim_{n \to +\infty} n^2 + n = +\infty $$ $$ \lim_{n \to +\infty} 3^n = +\infty $$ $$ \lim_{n \to +\infty} \left(\frac{2}{5}\right)^n = 0 $$ $$ \lim_{n \to +\infty} \frac{n^2 + 4n}{n^2 + 5} = 1 $$ $$ \lim_{n \to +\infty} \frac{1 + 4n}{3n + 6} = \frac{4}{3} $$ $$ \lim_{n \to +\infty} 2n^3 = +\infty $$