Here are the solutions to the problems: Question 8: How many ways can you arrange the letters of the word "COMBINATION"? The word "COMBINATION" has 11 letters. Count the occurrences of each letter: C: 1 O: 2 M: 1 B: 1 I: 2 N: 2 A: 1 T: 1 This is a permutation with repetitions. The formula is $\frac{n!}{n_1! n_2! ... n_k!}$, where $n$ is the total number of letters and $n_i$ are the counts of repeated letters. Step 1: Identify the total number of letters and the counts of repeated letters. Total letters $n = 11$. Repeated letters: O (2 times), I (2 times), N (2 times). Step 2: Apply the permutation with repetitions formula. $$ \frac{11!}{2! \times 2! \times 2!} = \frac{39916800}{2 \times 2 \times 2} = \frac{39916800}{8} = 4989600 $$ The calculated answer is 4,989,600. This is not among options a), b), c), or d). The final answer is $\boxed{\text{None of the above}}$. Question 9: In how many ways can you arrange the letters of the word "SUCCESS"? The word "SUCCESS" has 7 letters. Count the occurrences of each letter: S: 3 U: 1 C: 2 E: 1 This is a permutation with repetitions. Step 1: Identify the total number of letters and the counts of repeated letters. Total letters $n = 7$. Repeated letters: S (3 times), C (2 times). Step 2: Apply the permutation with repetitions formula. $$ \frac{7!}{3! \times 2!} = \frac{5040}{(3 \times 2 \times 1) \times (2 \times 1)} = \frac{5040}{6 \times 2} = \frac{5040}{12} = 420 $$ The calculated answer is 420. This is not among options a), b), c), or d). The final answer is $\boxed{\text{None of the above}}$. Question 10: If there are 5 men and 4 women, how many ways can a committee of 3 men and 2 women be formed? This is a combination problem, as the order of selection for committee members does not matter. We need to choose 3 men from 5 men AND 2 women from 4 women. Step 1: Calculate the number of ways to choose 3 men from 5. $$ \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10 $$ Step 2: Calculate the number of ways to choose 2 women from 4. $$ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3}{2 \times 1} = 6 $$ Step 3: Multiply the results from Step 1 and Step 2 to find the total number of ways to form the committee. $$ 10 \times 6 = 60 $$ The calculated answer is 60. This is not among options a), b), c), or d). The final answer is $\boxed{\text{None of the above}}$. Question 11: A bag contains 6 red balls and 4 green balls. In how many ways can 3 balls be drawn without replacement? Total number of balls = $6 \text{ (red)} + 4 \text{ (green)} = 10$ balls. We need to choose 3 balls from these 10. Since the order of drawing does not matter, this is a combination problem. Step 1: Use the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. Here, $n=10$ and $k=3$. $$ \binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10!}{3!7!} $$ Step 2: Calculate the value. $$ \binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 10 \times 3 \times 4 = 120 $$ The correct option is b). The final answer is $\boxed{\text{120}}$. Question 12: How many different