Building a telephone number with special digits [Combinatorics]

Question

It's a bit easy question and I should feel ashamed for asking it, but I really don't know how to solve it. Consider a telephone number with the size of $7$. How many different telephone numbers that starts with the number $2$, can we build with the following numbers: $1,2,2,2,3,3,3$.

I understand that there is only one option for the first number (should be $2$). But how can I arrange the other numbers? I though of using the Multinomial coefficients, but I'm not sure.

How can I solve this issue?

Answer 1

Once the first number is a $2$, we essentially have to figure out the remaining $6$ numbers out of the multiset $\{1,2,2,3,3,3\}$. Your intuition is correct that you can use multinomial coefficients, but let's demystify that:

We have $\binom{6}{3}$ ways of choosing where the $3$'s go; from there, we just have to choose where the place the two remaining $2$'s in the remaining three spots, so we have to multiply by $\binom{3}{2}$. If we'd like, we can also multiply by $\binom{1}{1}$, the number of ways to place the $1$. Then we get $$\binom{6}{3}\binom{3}{2} = \binom{6}{1,2,3}\,.$$

Answer 2

I'm going to phrase this in terms of symbols in a string, which you can convert to a more general problem.

Suppose you have $k$ symbols $a_1,a_2,...a_k$, such that there is an $\alpha_j$ number of $a_j$. Such that there are $A=\sum_{j=1}^k \alpha_j$ total characters in your string. The number of possible strings containing a $\alpha_1$ number of $a_1$ and an $\alpha_2$ number of $a_2$, etc, is $$\frac{A!}{\alpha_1! \alpha_2!...\alpha_k!}.$$

In your case we fix the first two and we wish to find the number of strings composed of 1, 2, 2, 3, 3, 3. This is $=\frac{6!}{1! \cdot 2! \cdot 3!}$.