We can start by figuring out the total number of permutations of the 10 digits, which is:
10! / (2! x 2! x 3!)
This is because there are 2 copies of the digit 1, 2 copies of the digit 2, 2 copies of the digit 3, and 1 copy each of the digits 4, 5, and 6. So we need to divide by the factorials of the numbers of copies of each digit.
Simplifying the expression, we get:
10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 / (2 x 1 x 2 x 1 x 3 x 2 x 1)
= 1,890,000
Therefore, there are 1,890,000 ways to arrange the numbers to form a 10-digit number.
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Answer:
We can start by figuring out the total number of permutations of the 10 digits, which is:
10! / (2! x 2! x 3!)
This is because there are 2 copies of the digit 1, 2 copies of the digit 2, 2 copies of the digit 3, and 1 copy each of the digits 4, 5, and 6. So we need to divide by the factorials of the numbers of copies of each digit.
Simplifying the expression, we get:
10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 / (2 x 1 x 2 x 1 x 3 x 2 x 1)
= 1,890,000
Therefore, there are 1,890,000 ways to arrange the numbers to form a 10-digit number.
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