Given word is INEONVENIENCE. The distinet letters are I, N, C, O, V, E (6 letters) and their frequencies are
I−2N−4C−2O−1V−1E−3Total number of 5-letters words
(i) When 5 distinct letters, number of words
=6P5=720(ii) One pair and 3 distinct letters,
number of words =4C1×5C3×=4×10×60=2400(iii) One triple and 2 distinct letters, number of words
=2C1×5C2×=2×10×20=400 (iv) Two pairs and one distinct, number
of words =4C2×4C1×=6×4×30=720(v) One triplet and one pair, number of words
=4×=4×10=40(vi) One 4 a like and one different
=1C1×5C1×=25∴ Number of words with at least one repealed letters
=(720+2400+400+720+40+25)−720=3585
