Distinct sequences of characters

Question

I'm new to combinatorial calculus. I have a question for you that I am trying to answer.

How many distinct sequences can be formed using exactly 5 times the letter A, 6 times the letter B and 7 times the letter C?
How many can be formed by choosing an additional letter of the alphabet from the remaining 23?

I asked the first question with the following calculation: $(18!)/(5!6!7!) $ (sorry for the fraction but I have not found a way to write it correctly).

For the second question, I have my doubts as to how to resolve it. Do I have to approach this question thinking that I only have one more place in the sequence, or do I have to understand how many ways I can occupy one of the 19 places by selecting a letter from 23?

Answer 1

First answer is correct and is the multinomial coefficient $\binom{18}{5,6,7}=\dfrac{18!}{5!6!7!}$.

The second answer can be found a number of ways, most introductory ways will involve multiplication principle and binomial or multinomial coefficients in some way.

As alluded to in comments above, one way would be to arrange five $A$'s, six $B$'s, seven $C$'s, and one $\heartsuit$ into a line and then pick which letter the $\heartsuit$ is replaced with.

$\binom{19}{5,6,7,1}\cdot 23=\frac{19!}{5!6!7!1!}\cdot 23$

This is equivalent to first arranging five $A$'s, six $B$'s, seven $C$'s in a line and then picking a spot to insert a $\heartsuit$ inbetween two characters or the front or back and then choosing which letter $\heartsuit$ represents.

$\binom{18}{5,6,7}\cdot 19\cdot 23$

You could completely reverse the order of steps as well in this case by first picking what the new letter is, then picking where it goes, then picking where the $A$'s go, then where the $B$'s, finally the $C$'s

$23\cdot 19\cdot \binom{18}{5}\cdot \binom{13}{6}\cdot 1$

In the end, just check that each outcome you wish to count is counted, and further each outcome is counted only once and not multiple times each.