How many numbers of different 4 digits can be formed by using the digits \( 1,2,0,4,5,7,3 \) ? A- \( { }^{7} p_{4}-{ }^{6} p_{3} \) B- \( \quad 4!-3 \) ! C- \( { }^{7} C_{4} \) D- \( { }^{7} p_{4} \)

To form a 4-digit number using the digits \( 1, 2, 0, 4, 5, 7, 3 \), we have to ensure that the first digit is not zero (0), as that would make it a 3-digit number. First, we can select the first digit from the digits \( 1, 2, 4, 5, 7, 3 \), which gives us 6 options. After choosing the first digit, we have 6 remaining digits including 0, which can fill the other three positions. Therefore, if we select any of the initial non-zero digits, the total number of combinations can be calculated as follows: 1. Choose the first digit: 6 options 2. Choose the next 3 digits from the remaining 6 digits (including 0, since it won't be the first digit): This can be done in \( {}^{6} P_{3} \) ways. Putting that all together, the total number of valid 4-digit combinations is \( 6 \times {}^{6} P_{3} \). So the correct expression does not match any of the options provided, but one could express the entire calculation as \( 6 \times \left( \frac{6!}{(6-3)!} \right) \). Now, let’s explore two fun related insights: Some of the digits provided have historical mathematical significance! For instance, the digit 0 is not just a placeholder; it revolutionized mathematics by enabling the development of algebra, calculus, and even computer science – how cool is that! And did you know that there are countless games and puzzles using permutations of numbers? You can try making numbers with dice rolls or even with everyday objects like blocks. It’s a great way to explore combinations and permutations practically while having fun!