# Math Help - ways

1. ## ways

How many ways are there to arrange 4 americans, 3 russians and 5 chinese into a queue, in such a way that no nationality forms a single consecutive block?

2. (12C4)(8C3)(5C5)-(9C1+10C1+8C1)
=166320-27
=166293

I not sure....

3. Originally Posted by asc
How many ways are there to arrange 4 americans, 3 russians and 5 chinese into a queue, in such a way that no nationality forms a single consecutive block?
Complementary Counting

I will assume the people are not identical

Find the number of ways in which they do form a consecutive block and then you can just subtract this number from the total.

I think it will run into a few cases:
Total number of ways in which americans form a block = 9!
Total number of ways in which russians form a block = 10!
Total number of ways in which chinese form a block = 8!

But we have counted some solutions multiple number of times.

Total number of ways in which russians and americans form a block = 7!
Total number of ways in which russians and chinese form a block = 6!
Total number of ways in which americans and chinese form a block = 5!
Total number of ways in which russians, americans and chinese form a block = 3!

So total number of ways: 9!+10!+8! - (7!+6!+5!) + 3! (You can also say ,"by principle of inclusion and exclusion")
Now subtract it from 12!
So
Answer:12! - (9!+10!+8! - (7!+6!+5!) + 3!)
I got 474975474 ways

P.S:I am most likely wrong. I find counting questions hard always. But I have tried working this problem. Tell me if you did not understand something, most likely the step is wrong

4. Originally Posted by Isomorphism
Total number of ways in which americans form a block = 9!
Total number of ways in which russians form a block = 10!
Total number of ways in which chinese form a block = 8!
They must be 9! 4!, 10! 3! and 8! 5!. You took americans as a group but probably forgot that there are 4! american groups. Same goes for all.

5. I do not think that either of the two replies is correct.
Let A be the event that all the Americans are together as a block. Then use R and C stand for the same for the other two groups. Now we want the number $\left| {A^c R^c C^c } \right|$, that is not A, not R and not C (none of the groups forms a single consecutive block).

You will want to justify each of the following.
The number of ways that all the American forms a single consecutive block is $\left| A \right| = \left( {4!} \right)\left( {9!} \right)$, WHY?.
The number of ways that all the American and all of the Russians are together as a consecutive block is $\left| AR \right| = \left( {4!} \right) \left( {3!} \right)\left( {7!} \right)$, WHY?.

The number of ways that at least one of the national groups forms a single consecutive block is:
$\left| {A \cup R \cup C} \right| = \left| A \right| + \left| B \right| + \left| C \right| - \left| {AR} \right| - \left| {AC} \right| - \left| {RC} \right| + \left| {ARC} \right|$.

Now you want $\left| {A^c R^c C^c } \right| = 12! - \left| {A \cup R \cup C} \right|$.