- Suppose a word of n letters has n1 of one kind of letter, n2 of a second kind, n3 of a third kind, and so on with n=n1+n2+n3+...+nk then the number of distinguishable permutations of the n letters is =
n!
n1!×n2!×n3!...nk!
We have, word 'AGAIN' Arrange alphabetically: AAGIN The letter starts from A, A A G I N AAGIN = 4! = 24 The letter starts from G, G A A I N GAAIN = (4!)/2 = 12 The letters starts from I, I A A G N IAAGN = (4!)/2 = 12 Number of total letters = 24 + 12 + 12 = 48 49th word is NAAGI 50th word is NAAIG.