How many four letter words can be formed using the letters of the word 'INEFFECTIVE' so that
(i)3 alike letters and one distinct letter
(ii)2 alike letters of one kind & 2 distinct letters
(iii)all distinct letters
are there

I) The only possible three repeated letters are Es, because only E is repeated thrice. The choice of the fourth letter is from I, N, F, C, T and V (6 choices).

Total number of permutations of these is hence 6C1 x (4!) / (3!) = 24.

II) There are 3 possibilities for the 2 alike letters: either II, EE or FF. If you choose I as your repeated letter, any 2 letters from N, E, F, C, T and V will be your choices for the two distinct letters. If you choose E as your repeated letter, any two letters from N, I, F, C, T and V will be your choices for the two distinct letters. If you choose F as your repeated letter, any two letters from E, N, I , C, T and V will be your choices for the two distinct letters.

Total number of permutations is hence 3C1 x 6C2 x (4!) / (2!) = 540

III) All your four distinct letters will be from I, N, E, F, C, T and V.

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TopNewestI) The only possible three repeated letters are Es, because only E is repeated thrice. The choice of the fourth letter is from I, N, F, C, T and V (6 choices).

Total number of permutations of these is hence 6C1 x (4!) / (3!) = 24.

II) There are 3 possibilities for the 2 alike letters: either II, EE or FF. If you choose I as your repeated letter, any 2 letters from N, E, F, C, T and V will be your choices for the two distinct letters. If you choose E as your repeated letter, any two letters from N, I, F, C, T and V will be your choices for the two distinct letters. If you choose F as your repeated letter, any two letters from E, N, I , C, T and V will be your choices for the two distinct letters.

Total number of permutations is hence 3C1 x 6C2 x (4!) / (2!) = 540

III) All your four distinct letters will be from I, N, E, F, C, T and V.

Total number of permutations is hence 6P4 = 360

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