Problema Solution
Subject identification numbers in a certain scientific research project consist of twotwo digitsdigits followed by twotwo lettersletters and then threethree more digitsdigits. Assume repetitions are not allowed within any of the three groups, but digitsdigits in the first group of twotwo may occur also in the last group of threethree. How many distinct identification numbers are possible?
Answer provided by our tutors
To solve this problem we will use permutations without repetition.
Assuming the digits are chosen from the set of 10 elements {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} and the letters are chosen from the set of alphabet letters of 26 elements.
P(10, 2)*P(26, 2)*P(10, 3) = (10!/(10 - 2)!)(26!/(26 - 2)!)(10!/(10 - 3)!) = (10!*26!*10!)/(8!24!7!) =
= 9*10*25*26*8*9*10 = 42,120,000
42,120,000 distinct identification numbers are possible.