Arranging $3$ groups of people

Question

We have $4$ Americans, $3$ Russians and $5$ Chinese. They are forming a line and no nationality is allowed to form a block. How many possibilities are there?

A block just means that something like $AA...$ isn't allowed.

We have $12!/4!3!5!$ possibilities if we allow blocks, now we have to substract the possibilities with blocks. Let's take $2A$ and force them to be next to each other, we get $11$ possibilities; so $11(10!/2!3!5!)$. We have to do the same with $R$ and $C$ and we get

$12!/4!3!5! - 11(10!/2!3!5! + 10!/4!1!5! + 10!/4!3!3!)$. Is this correct?

I just used the $n!/k_1!k_2!...$ formula and don't really know where it comes from, can someone explain that maybe? And is there a more elegant solution?

I just realized that my solution is negative, that's very unfortunate — Philip730, Aug 11 '22 at 00:43
$(k_1+k_2+\cdots k_n)!/ k_1!k_2!\cdots k_n!$ is the multinomial coefficient, and counts distinguishable ways to arrange a multiset of $k_1$ identical items of type 1, $k_2$ identical items of type 2, et cetera. — Graham Kemp, Aug 11 '22 at 00:43
I would look at it recursively, keeping track of which nationality starts. It's easy with small groups and larger groups are made out of smaller ones. — lulu, Aug 11 '22 at 00:45
So do just two of the same nationality side by side form a block? Or must it be all of them? — Parcly Taxel, Aug 11 '22 at 01:10
The length of the block doesn't matter, $...AA...$ and $...AAA...$ aren't allowed — Philip730, Aug 11 '22 at 01:18
Maybe the answer to this question can guide you in the right direction? — Bergson, Aug 11 '22 at 15:25

true blue anil · Accepted Answer · 2022-08-11T04:36:49.667

I am afraid that with conventional methods, it will be a very tortuous affair, so I shall use the formula by Jair Taylor

Define polynimomials $q_k(x) = \Large\sum_{i=1}^k \frac{(-1)^{i-k}}{i!} {k-1 \choose i-1}x^i$ for $k\geq 1$, and $q_0(x) =1,$ e.g. for $k=2,q_2(x)$ works out to $(x2−2x)/2!$ The number of permutations will be given by

$$\int_0^\infty \prod_j q_{k_j}(x)\, e^{-x}\,dx.$$

The specific formula for this problem can be seen at Wolframalpha yielding the answer as $588$

This, of course, is if we don't care which particular persons are sitting next to each other, we are concerned only about two people of the same nationality not being together, as seems to be the intention.

Arranging $3$ groups of people

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