I am having a lapse in memory. I know I know this, but I don't recall how to calculate the number of permutations there are for a multiset.

Here is the actual problem and how I am going about solving it. I have a deck of 60 cards. 22 of them are of type A, 4 are of type B, 4 of type C, and the rest of type D. I want to know the probability that if I draw 9 cards, I get exactly 3 of A, 2 of B, 1 of C, and 3 of D where the 9th card chosen is not B. I know a long, drawn-out process to calculate it, but I thought there was an easier way. Here is my solution:

Suppose we have a sixty-element multiset with the following repetition numbers:

$$\{22\cdot A, 4\cdot B, 4\cdot C, 30\cdot D\}$$

And we want to know the number of 9-element permutations (not necessarily distinct). I can calculate this by figure out the number of permutations like this:

$$\sum_{b=0}^4 \sum_{c=0}^4 \sum_{a=0}^{9-b-c}\dbinom{22}{a}\dbinom{4}{b}\dbinom{4}{c}\dbinom{30}{9-a-b-c} \dfrac{9!}{a!b!c!(9-a-b-c)!}$$

This is the total number of possible ways to draw the first nine cards. Next, let's figure out the total number of ways to draw the first nine cards containing the elements I want:

$$\dbinom{22}{3}\dbinom{4}{2}\dbinom{4}{1}\dbinom{30}{3}\left(\dfrac{9!}{3!2!1!3!}-\dfrac{8!}{3!1!1!3!}\right)$$

I think this gives the correct answer, but it is not easy to figure out.

Here is the actual problem and how I am going about solving it. I have a deck of 60 cards. 22 of them are of type A, 4 are of type B, 4 of type C, and the rest of type D. I want to know the probability that if I draw 9 cards, I get exactly 3 of A, 2 of B, 1 of C, and 3 of D where the 9th card chosen is not B. I know a long, drawn-out process to calculate it, but I thought there was an easier way. Here is my solution:

Suppose we have a sixty-element multiset with the following repetition numbers:

$$\{22\cdot A, 4\cdot B, 4\cdot C, 30\cdot D\}$$

And we want to know the number of 9-element permutations (not necessarily distinct). I can calculate this by figure out the number of permutations like this:

$$\sum_{b=0}^4 \sum_{c=0}^4 \sum_{a=0}^{9-b-c}\dbinom{22}{a}\dbinom{4}{b}\dbinom{4}{c}\dbinom{30}{9-a-b-c} \dfrac{9!}{a!b!c!(9-a-b-c)!}$$

This is the total number of possible ways to draw the first nine cards. Next, let's figure out the total number of ways to draw the first nine cards containing the elements I want:

$$\dbinom{22}{3}\dbinom{4}{2}\dbinom{4}{1}\dbinom{30}{3}\left(\dfrac{9!}{3!2!1!3!}-\dfrac{8!}{3!1!1!3!}\right)$$

I think this gives the correct answer, but it is not easy to figure out.

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