Problem
Find the number of 4-digit integers in the closed interval [2022, 4482] formed by using the digits 0, 2, 3, 4, 6, 7.
Solution
To find the number of such integers, we’ll first count the total number of integers starting with either 2, 3, or 4, and whose remaining digits are from the given set.
Then, we’ll remove the integers that do not lie in the given interval, i.e. those less than 2022 and those greater than 4482.
The total number will be 3 x 6 x 6 x 6 or 648, since there are 3 choices for the first digit, and 6 choices each for the remaining three.
Now, we’ll remove the ‘extra’ integers, i.e. those lying outside the given interval.
The integers that are less than 2022 are of the form 200X and 202Y, where X can be any of the given digits and Y is less than 2.
For 200X, there are 6 possibilities, and for 202Y, there’s only 1, when Y equals 0. So, a total of 7 such integers.
The integers that are greater than 4482 must be of the form 4ABC, where A is greater than 4, and B and C can take any value from the given set of digits.
Note that we ignored numbers of the form 448X, because 8 isn’t a member of the given set of digits.
That is, 4483, 4484, 4486, and 4487 can’t be formed using the given digits, even though they’re greater than 4482.
So, given the above conditions, there will be 2 x 6 x 6 or 72 integers that are greater than 4482.
Finally, the required number of integers in the given interval will be 648 – 7 – 72 or 569.
Comments
A moderate problem on permutations, that involved counting of the complementary set. The main effort lied in carefully eliminating all the ‘extra’ integers from the total number of integers. What if the digits weren’t allowed to be repeated? Can you find out the number of integers in that case?