P R A C T I C E S
This word contains 8 distinct letters ( P, R, A, C, T, I, E, S), among them C is repeated twice (2).
The qs. is to make 3-letter words out of this letter.
So, we can form 3-letter words in this ways:
- All the letters of the word (3-letter word) will be Distinct.
OR
2. The word will be having a letter which is repeated Twice.
So, the 1st one:
There are 8 distinct letters, out of which we have to choose 3 letters and arrange them.
Hence, 8P3 = 8!/(8-3)! = 8!/5! = 336
2nd:
So, we're having only 1 letter which is repeated Twice (1C1) & we have to select another letter from remaining 7 letters (7C1) & then we have to arrange the word (3!/2!).
(here 1 thing may comes into mind that why 3!/2! ? : the 3 -letter word is containing 3 letters, out of which 1 letter is distinct and another 2 letter is identical(alphabet 'C'), so 1 letter is repeated twice in a 3 letter word, therefore 3!/2!)
Hence, 1C1 * 7C1 * 3!/2! = 1 * 7!/{1!*(7-1)!} * 3 = 7*3 = 21
Total ways = 1st one OR 2nd one
= 336 + 21
= 357