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Best answer
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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:

  1.  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  

        

selected by
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1 votes
$PRA'CC'TIES$
No of words with 0 repetition $: 8_{c_3}*3!=336$
No of words with 1 repetition $: 7*3=21 \\ Total : \color{Blue}{357}$
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In the given word we have two letters which are repeating and other seven letters that are not repeating.So make them partition like this (CC) PRATIES .Now two cases can be formed  first case where we will be taking one letter from the repeated letter's group and the other letters from the non-repeated ones and the second case where two letters will be taken from the repeated and the other one from the non-repeated.

Therefore,
1st case- Taking one C and other two from the non-repeated group.So total 8 letters.And from this 8 letters we have to select three to form three letter word.So 8C3 and their arrangement 3! i.e. 8C3*3!=336

2nd case- Taking two C's and other one letter from the non-repeated group.So from 1 group we have to select 1 which is 1C1 then remaining one letter from non-repeated letter so 7C1 and then their arrangements which is 3! but we have to divide the whole result with 2! because two C's are present.Thus it forms:  (1C1*7C1)*3!/2!  =21

Now either 1st case will be formed or second one so:336+21=357

Ans=357

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