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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, ^{8}P_{3 }= 8!/(8-3)! = 8!/5! = 336

**2nd:**

So, we're having only 1 letter which is repeated Twice (** ^{1}C_{1}**)

(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, ^{1}C_{1 }* ^{7}C_{1 }* 3!/2! = 1 * 7!/{1!*(7-1)!} * 3 = 7*3 = 21

**Total ways** = 1st one **OR** 2nd one

= 336 + 21

= **357 **

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