A database uses 20-character strings as encrypted unique identifiers. The valid characters in these strings are upper-case letters of the English alphabet and decimal digits. How many valid identifiers are possible if a valid identifier must meet all of the following criteria:

(a) Letter(s) from the set {A, E,I, O,U} occur in exactly three positions of the string.
(b) The last three characters in the string are distinct decimal digits that do not appear elsewhere in the string.
(c) The remaining characters of the string may be filled with any of the remaining letters or decimal digits.

Step-by-step explanation:

So we need to form various permutations and combinations for a 20 character string..

first we'll fill the last three places with distinct decimal digits as this is the only points that decides the order.

there are P(10,3) ways to fill these places

P(10,3) = 10*9*8 ways   P stands for permutation

Now we'll focus on the vowels {A,E,I,O,U}

The vowels occur in exactly three positions and we have 17 places left to fill.

So we need to choose the three places out of the 17 places and this can be done in C(17,3) ways.        

C is for combination

C(17,3) = 680 ways

Now there are 5^3 way to fill the three positions with the vowels

so for the vowels total ways = 680*5^3 = 85000

Now w are left with 14 places to fill and a total of 7 digits and 21 alphabets are lets as we can't use the vowels , so a total of 28 symbols are left.

And the symbols can be repeated

so total ways will be = 28^14 ways

hence the total number of ways for setting up the 20 character string is = 10*9*8 *680*5^3*28^14 ways

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