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.