Multiple choice

The number of ways of arranging the letters $AAAAA, BBB, CCC< D, EE$ and $F$ in a row if the letters $C$ are separated from one another is:

  1. ${ _{ }^{ 13 }{ C } }_{ 3 }\cfrac { 12! }{ 5!3!2! } $
  2. $\cfrac{13!}{5!3!3!2!}$
  3. $\cfrac{14!}{5!3!2!}$
  4. $\cfrac{15!}{3!{(3!)}^{2}!}-\cfrac{13!}{5!3!2!}-\cfrac{12!}{5!3!}{ _{ }^{ 13 }{ C } }_{ 2 }$
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A Correct answer
Explanation

Arrange the non-C letters first, giving 12!/(5!3!2!) arrangements. These letters create 13 gaps, and choosing 3 distinct gaps for the separated C letters gives C(13,3) × 12!/(5!3!2!).

AI explanation

The total number of letters is 15, which includes 5 A's, 3 B's, 3 C's, 1 D, 2 E's and 1 F. We first arrange the 12 non-C letters, which can be done in 12 factorial divided by (5 factorial times 3 factorial times 2 factorial) ways. This arrangement creates 13 gaps (including the ends) where the 3 C's can be placed to ensure they are separated. The number of ways to choose 3 gaps out of 13 is 13 C 3, making the total number of arrangements 13 C 3 multiplied by 12 factorial divided by (5 factorial times 3 factorial times 2 factorial).

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