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Answer :
Final answer:
The total number of natural numbers that can be made with the digits 3, 4, 5, and 6 in the range 10000 to 99999 such that each digit appears at least once is 89856.
Explanation:
In order to find the number of natural numbers that can be made with the digits 3, 4, 5, and 6 in the range 10000 to 99999 such that each digit appears at least once, we can use the concept of permutations. Firstly, we can calculate the total number of five-digit numbers in this range, which is 90000 (99999 - 10000 + 1 = 90000). Then, we need to subtract the number of five-digit numbers that do not include all four digits. We can determine this by counting the number of choices for each digit (4 choices) and taking this to the power of the number of digits (5 in this case). Therefore, the total number of natural numbers that meet the given condition is 90000 -[tex](4^5)[/tex]= 89856.
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