Multiple choice

The different number of words formed by using the letters $4A's,2C's,1H,1D,1B$ if both $C's$, do not occur together are

  1. $7560$
  2. $1680$
  3. $60$
  4. $5880$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Total letters = 9. Total arrangements = 9! / (4! * 2!) = 7560. Arrangements where 2 C's are together: Treat CC as one unit, total units = 8. Arrangements = 8! / 4! = 1680. Arrangements where C's are not together = 7560 - 1680 = 5880.

AI explanation

Using the method of total arrangements minus restricted arrangements, first find the total arrangements of the 9 letters (4A, 2C, H, D, B) as 9! / (4! * 2!) = 7560. Treating the two C's as a single block gives 8! / 4! = 1680 arrangements where they are together. Subtracting the restricted arrangements from the total gives 7560 minus 1680, which equals 5880.

Keep practicing — related questions