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Get access to the detailed solutions to the previous years questions asked in IIFT exam
There are 48 alphabets out of which we need 3
So, no. of ways of selecting 3 alphabets from 48 alphabets is 48C3
Since we need to arrange these alphabets in ascending order so there is only one possible way for every three alphabets.
The first digit is 3 more than the third digit.
For a particular first digit there will be a particular third digit but the 2nd digit can be any of the numbers
So, no. of ways of selecting first digit= 15
No. of ways of selecting 2nd digit = 18
Therefore, total no. of ways = = 4669920