How many six digit number can be formed by the digits 2,2,3,3,3,4 by permutation 

To find how many six-digit numbers can be formed with the digits \(2, 2, 3, 3, 3, 4\), we can use the formula for permutations of multiset:

\[
\text{Number of permutations} = \frac{n!}{n_1! \cdot n_2! \cdot n_3!}
\]

where:
- \(n\) is the total number of digits,
- \(n_1, n_2, n_3\) are the frequencies of each distinct digit.

Here, we have:
- Total digits \(n = 6\)
- The digit \(2\) appears \(2\) times,
- The digit \(3\) appears \(3\) times,
- The digit \(4\) appears \(1\) time.

Now, substituting into the formula:

\[
\text{Number of permutations} = \frac{6!}{2! \cdot 3! \cdot 1!}
\]

Calculating each factorial:
- \(6! = 720\)
- \(2! = 2\)
- \(3! = 6\)
- \(1! = 1\)

Thus, we have:

\[
\text{Number of permutations} = \frac{720}{2 \cdot 6 \cdot 1} = \frac{720}{12} = 60
\]

So, **60 different six-digit numbers can be formed** using the digits \(2, 2, 3, 3, 3, 4\).

Would you like more details on this calculation, or do you have any questions? Here are some related questions you might consider:

1. How would the number of permutations change if one of the digits were different?
2. What is the total number of distinct arrangements for a set of digits with no repetitions?
3. Can you provide examples of the six-digit numbers formed with these digits?
4. How do you calculate permutations for a larger set of digits?
5. What other mathematical principles are related to permutations? 

**Tip:** When calculating permutations, always account for repeated elements to avoid overcounting.

How many six digit number can be formed by the digits 2,2,3,3,3,4 by permutation

Discover how to calculate the number of six-digit numbers that can be formed using the digits 2, 2, 3, 3, 3, 4. This guide covers the permutations of multisets, providing a clear solution with detailed calculations. Suitable for students in Grades 7-9.

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