UPSC NDA II 2015 Mathematics Solved Paper

- 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.