Counting Numbers Between 0 and 9999 with Exactly One Digit 7 and One Digit 2
brIn this article, we delve into the realm of combinatorial mathematics to explore an interesting problem: how many numbers between 0 and 9999 have exactly one digit 7 and exactly one digit 2. This exploration will not only provide a solution but also highlight the techniques and principles used in such counting problems.
Introduction:
The challenge at hand requires us to find numbers with specific digit placement. We begin by understanding the range of numbers in question, and then proceed with a systematic approach to determine the number of possibilities.
Total Number of Digits:
We are considering all numbers from 0000 to 9999, which translates to a total of 5000 numbers (including 0000).
Placement of the Digits:
The problem demands that each number contains exactly one digit 7 and exactly one digit 2. To start, we need to determine the number of ways to place these digits in the four positions available.
Since the numbers can have up to 4 digits, we have 4 positions to choose from. We need to select 2 out of these 4 positions for the digits 7 and 2. The number of ways to choose 2 positions out of 4 is given by the binomial coefficient:
Binomial coefficient: C ( n , r ) ( n ! / r ! ? ( n - r ) ! )
In our case, n 4 (total positions) and r 2 (positions to fill with 7 and 2), so:
C ( 4 , 2 ) 4 ! / ( 2 ! ? ( 4 - 2 ) ! ) 6
However, since the order of the positions matters (7 can be in the first chosen position and 2 in the second, or vice versa), we multiply by 2:
Total ways to choose positions: 6 times; 2 12
Filling the Remaining Positions:
Once we have placed the digits 7 and 2, the remaining 2 positions can be filled with any of the 8 digits (0, 1, 3, 4, 5, 6, 8, 9). For each of these remaining positions, we have 8 choices.
Total ways to fill remaining positions: 8 times; 8 64
Total Combinations:
To find the total number of numbers that meet the criteria, we multiply the number of ways to choose and arrange 7 and 2 by the number of ways to fill the remaining positions:
Total combinations: 12 times; 64 768
Conclusion:
Therefore, there are 768 numbers between 0 and 9999 that have exactly one digit 7 and exactly one digit 2. This problem showcases the power of combinatorial mathematics in solving intricate numerical challenges.
Auxiliary Content for Fun:
Lets allow leading zeroes just for fun, as they don’t change the answer in this context. If you consider hexadecimal base 16 numbers, you have 4 choices of where to put the 7. This leaves three choices for where to place the 2. So, 4 times; 3 12. Each of the remaining two spots can have any of the 14 remaining hexadecimal digits (excluding 2 and 7). Hence:
Total 4 × 3 × 14 × 14 2352
Lastly, for decimal numbers, the calculation remains similar, showcasing how the principle of combinatorial counting applies universally.