how many 4-letter words can be formed using the letters of the word FAILURE so that

1) Fis included in each word 2) F is not included in any wrd

isn't the answer supposed to be 6P3? the answer given in the reference book is 6C3 X 4! =480

(1) F included in the 4-letter-word:-

There are four cases in which F is present in the word.

CASE 1 :- F is the first letter of the 4-letter word,

The rest 3 letters required can be chosen in ⁷C₃ ways from the word FAILURE. And these 3 letters can be arranged in 3!.

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No. of 4-letter words with F as the first letter =⁷C₃* 3!

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Case 2:-F is the second letter of the 4-letter word,

The rest 3 letters required can be chosen in⁷C₃ways from the word FAILURE. And these 3 letters can be arranged in 3!.

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No. of 4-letter words with F as the second letter =⁷C₃* 3!

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Similarily for Case 3 and Case 4 :-⁷C₃* 3! ways each.

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Hence totally if F is incuded then the no. of 4-letter words = 4 *⁷C₃* 3! = 4*35*6 = 840 words.

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(2) F is not included in the 4-letter word,

Then choosing 4 letters rest of the 6 letters can be done in ⁶C₃ ways and it can be arranged in 4! ways.

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Hence no. of 4-letter-words formed without F = ⁶C₃ * 4! = 480 words.

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