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Question 8: How many ways can you arrange the letters of the word "COMBINATION"?

Mathematics

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Answer

B

Step-by-step solution

9 steps

Here are the solutions to the problems:

Part 8· 2 steps

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 n!n1!n2!...nk!\frac{n!}{n_1! n_2! ... n_k!}, where nn is the total number of letters and nin_i are the counts of repeated letters.

  1. Identify the total number of letters and the counts of repeated letters.

    Total letters n=11n = 11.
    Repeated letters: O (2 times), I (2 times), N (2 times).

  2. Apply the permutation with repetitions formula.

    11!2!×2!×2!=399168002×2×2=399168008=4989600\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 None of the above\boxed{\text{None of the above}}.

Part 9· 2 steps

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.

  1. Identify the total number of letters and the counts of repeated letters.

    Total letters n=7n = 7.
    Repeated letters: S (3 times), C (2 times).

  2. Apply the permutation with repetitions formula.

    7!3!×2!=5040(3×2×1)×(2×1)=50406×2=504012=420\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 None of the above\boxed{\text{None of the above}}.

Part 10· 3 steps

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.

  1. Calculate the number of ways to choose 3 men from 5.

    (53)=5!3!(5−3)!=5!3!2!=5×42×1=10\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10

  2. Calculate the number of ways to choose 2 women from 4.

    (42)=4!2!(4−2)!=4!2!2!=4×32×1=6\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3}{2 \times 1} = 6

  3. Multiply the results from Step 1 and Step 2 to find the total number of ways to form the committee.

    10×6=6010 \times 6 = 60

    The calculated answer is 60. This is not among options a), b), c), or d).
    The final answer is None of the above\boxed{\text{None of the above}}.

Part 11· 2 steps

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 (red)+4 (green)=106 \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.

  1. Use the combination formula (nk)=n!k!(n−k)!\binom{n}{k} = \frac{n!}{k!(n-k)!}.

    Here, n=10n=10 and k=3k=3.

    (103)=10!3!(10−3)!=10!3!7!\binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10!}{3!7!}

  2. Calculate the value.

    (103)=10×9×83×2×1=10×3×4=120\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 120\boxed{\text{120}}.

Part 12

How many different

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