Give the total number of possible arrangements of 3 letters chosen from the word CALCULUS. The answer is 96, but all I can get is 5P3=60 (permutations of 3 from 5 different elements), or 8P3 adjusted for three pairs of identical elements, i.e. 8!/(2!2!2!5!)=42.
I would really appreciate the workings. Many thanks for your help.
Best Answer
Supposing you may not use any letter more times then present in CCLLUUAS, you arrangement either contains $3$ distinct letters or two equal letters and a third distinct letter (there are no letters present three times). In the former case you have $\binom53=10$ choices to the subset of letters and $6$ permutations of the letter, for $60$ possibilities. In the latter case you have $3$ choices for the doubled letter, $4$ choices for the remaining letter, and $3$ distinct permutations, for $36$ possibilities. In total $60+36=96$ possibilities.