Counting Four-Digit Integers with Exactly One 2 and One 5
In this article, we will explore how many four-digit positive integers have exactly one digit equal to 2 and exactly one digit equal to 5. This problem requires a careful breakdown into several steps, utilizing combinatorial methods to arrive at the final answer.
Step 1: Choosing the Positions for 2 and 5
The first step is to choose 2 out of the 4 digit positions for the digits 2 and 5. The number of ways to choose 2 positions from 4 is given by the combination formula (binom{n}{k}), which in this context is (binom{4}{2} 6).
Step 2: Assigning Digits 2 and 5
For each of the chosen positions, we can assign the digits 2 and 5 in two different ways: either 2 in the first chosen position and 5 in the second chosen position, or vice versa. Therefore, for each of the 6 combinations, there are 2 arrangements, resulting in a total of 12 ways to assign the digits 2 and 5.
Step 3: Choosing the Remaining Two Digits
The remaining two positions must be filled with digits that are neither 2 nor 5. There are 8 possible digits (0, 1, 3, 4, 6, 7, 8, and 9) to choose from. However, since we are forming a four-digit number, the first digit cannot be 0.
Case 1: If one of the remaining positions is the first digit and it cannot be 0, we have 7 choices (1, 3, 4, 6, 7, 8, 9).
For the other remaining position, we can use any of the 8 digits including 0. Therefore, the number of ways to fill the remaining positions in this case is:
7 ways (first position) × 8 ways (second position) 56 ways.
Case 2: If the first position is filled with one of the digits 2 or 5 (which we have already placed in the first chosen position), this case cannot apply here because the 2 and 5 have been fixed in the chosen positions.
Step 4: Total Count
To find the total count, we multiply the number of ways to arrange 2 and 5 by the number of ways to fill the remaining two digits:
12 ways to arrange 2 and 5 × 56 ways to fill other digits 672.
Thus, the total number of four-digit positive integers that have exactly one digit equal to 2 and exactly one digit equal to 5 is 672.
Alternative Solution
Alternatively, we can use a complementary counting approach by breaking the problem into two cases:
Case 1: All 4-Digit Numbers that Work (including leading 0)
In this case, we first pick the two positions for 2 and 5 and then where the other two digits go. The number of such cases is 6!
Case 2: All Possible 4-Digit Numbers with Leading 0
Here, we pick two out of the remaining two spots of the leading zero four-digit number to have 2 and 5, and the last digit has 8 other choices. Number of such cases is 3!
The final answer is obtained by subtracting Case 2 from Case 1:
6! - 3! × 8 720 - 48 720.
This method yields 720, which is one more than the previous method due to the inclusion of leading zeros, but the primary target of the question is positive four-digit integers.
By thoroughly examining both methods, we can understand the nuances of combinatorial counting and see how each method arrives at the correct total number of four-digit integers that meet the criteria.