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How many numbers of 6 digits can be formed with the digits 2, 3, 2, 0, 3, 3?

Question: How many numbers of 6 digits can be formed with the digits 2, 3, 2, 0, 3, 3?

Solution:
The digits 2, 3, 2, 0, 3, and 3 can be arranged in 6! / (3! * 2!) = 720 / (2 * 6) = 60 ways.
However, arrangements that start with 0 are not valid 6-digit numbers, and there are $\frac{5!}{2! * 3!}$ = $\frac {120} {12}$ = 10 such arrangements that start with 0.
Therefore, the correct number of valid 6-digit numbers that can be formed from the digits 2, 3, 2, 0, 3, and 3 is 60 - 10 = 50.

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