How many five-digit positive integers comprising only the digits 1, 2, 3, and 4, each appearing at least once, exist such that the number is divisible by 4?

Question

Harsh Garg · Accepted Answer

We need to find the number of five-digit positive integers comprising only the digits 1, 2, 3, and 4, each appearing at least once, such that the number is divisible by 4.
A number is divisible by 4 if its last two digits form a number divisible by 4. Therefore, we need to ensure the last two digits are one of the following: 12, 24, 32, 44.
We can split the problem into cases based on the last two digits.
Case 1: Last two digits are 12The remaining digits are 1, 2, 3, and 4, each appearing at least once. We need to select 3 digits out of the 4 remaining digits to fill the first three positions.
The number of permutations of 3 out of 4 digits is P(4, 3) = 4! / (4 - 3)! = 24.
Case 2: Last two digits are 24The remaining digits are 1, 2, 3, and 4, each appearing at least once. We need to select 3 digits out of the 4 remaining digits to fill the first three positions.
The number of permutations of 3 out of 4 digits is P(4, 3) = 4! / (4 - 3)! = 24.
Case 3: Last two digits are 32The remaining digits are 1, 2, 3, and 4, each appearing at least once. We need to select 3 digits out of the 4 remaining digits to fill the first three positions.
The number of permutations of 3 out of 4 digits is P(4, 3) = 4! / (4 - 3)! = 24.
Case 4: Last two digits are 44The remaining digits are 1, 2, 3, and 4, each appearing at least once. We need to select 3 digits out of the 4 remaining digits to fill the first three positions.
The number of permutations of 3 out of 4 digits is P(4, 3) = 4! / (4 - 3)! = 24.
Summing all the cases:24 + 24 + 24 + 24 = 96
Thus, there are 96 five-digit positive integers comprising only the digits 1, 2, 3, and 4, each appearing at least once, that are divisible by 4.