5 Chinese, 5 Russians and 5 Mexicans are sitting in a row in such a way that the Chinese cannot be seated in the first 5 seats, the Russians cannot be seated in the 5 middle seats and the Mexicans cannot be seated in the last 5 seats. In how many different ways can they sit?

The only thing I can tell is that the total number of possible arrangements, without the restriction, is 15!/5!*5!*5! based on the 3rd theorem. But I don't know how to apply the restriction!

X

5 Chinese, 5 Russians and 5 Mexicans are sitting in a row in such a way that the Chinese cannot be seated in the first 5 seats, the Russians cannot be seated in the 5 middle seats and the Mexicans cannot be seated in the last 5 seats. In how many different ways can they sit?

Relevant page
<a href="/counting-probability/3-permutations.php">3. Permutations</a>
What I've done so far
The only thing I can tell is that the total number of possible arrangements, without the restriction, is 15!/5!*5!*5! based on the 3rd theorem. But I don't know how to apply the restriction!

Let the row of seats be divided into 3 segments, which I'll call "NC" (for non-Chinese, "NR" (for non-Russians) and "NM" (for non-Mexicans).

NC

NR

NM

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

The Chinese can sit in any NR or NM chairs, the Russians in either NC or NM chairs, and Mexicans in NC and NR.

We need to regard the 5 Chinese as 1 "unit", the 5 Russians as 1 "unit", and the 5 Mexicans as 1 "unit'.

For each segment, the remaining 10 people need to be thought of as 6 "units" (the 5 people from one country who can sit anywhere in the remaining 10 chairs, plus 1 "unit" which is people from the other country segment that can only sit in 5 of the remaining chairs.

I think I got all that right!

Try to continue on from there.

Regards
Murray

X

Hi Alex
Let me get you started.
Let the row of seats be divided into 3 segments, which I'll call "NC" (for non-Chinese, "NR" (for non-Russians) and "NM" (for non-Mexicans).
<table class="tableex"><tr><td>NC</td><td>NR</td><td>NM</td></tr><tr><td>1 2 3 4 5</td><td>1 2 3 4 5</td><td>1 2 3 4 5</td></tr></table>
The Chinese can sit in any NR or NM chairs, the Russians in either NC or NM chairs, and Mexicans in NC and NR.
We need to regard the 5 Chinese as 1 "unit", the 5 Russians as 1 "unit", and the 5 Mexicans as 1 "unit'.
For each segment, the remaining 10 people need to be thought of as 6 "units" (the 5 people from one country who can sit anywhere in the remaining 10 chairs, plus 1 "unit" which is people from the other country segment that can only sit in 5 of the remaining chairs.
I think I got all that right!
Try to continue on from there.
Regards
Murray

Alex did not respond. Let me have a go at the next step.

Let's consider the group of Chinese people first.

In the 5 seats in the NC segment, we can only have Russians or Mexicans. There are `(10!)/((10-5)!) = (10!)/(5!)` ways to arrange the Russians and Mexicans in this first segment of seats.

For the other 2 seat segments (NR and NM), the Chinese can sit anywhere in the 10 seats, so `(10!)/((10-5)!) = (10!)/(5!)` ways, the Russians can only sit in the NM seats (`5!` ways) and the Mexicans can only sit in the NR seats (also `5!` ways).

How am I going so far?

X

Murray
Alex did not respond. Let me have a go at the next step.
Let's consider the group of Chinese people first.
In the 5 seats in the NC segment, we can only have Russians or Mexicans. There are `(10!)/((10-5)!) = (10!)/(5!)` ways to arrange the Russians and Mexicans in this first segment of seats.
For the other 2 seat segments (NR and NM), the Chinese can sit anywhere in the 10 seats, so `(10!)/((10-5)!) = (10!)/(5!)` ways, the Russians can only sit in the NM seats (`5!` ways) and the Mexicans can only sit in the NR seats (also `5!` ways).
How am I going so far?

task 1: 5 seats in the NC segment can be occupied by 10 people
this can be done in 10x9x8x7x6=10!/5!(=30240) ways
task 2: 5 seats in the RC segment can be occupied by 10 people
this can be done in 10!/5! ways
task 3: 5 seats in the MC segment can be occupied by 10 people
this can be done in 10!/5! ways
now by applying multiplication principle of counting because actual task is achieved when all task 1 task 2 task 3 occur
so number of different ways they can sit= 10!/5! x 10!/5! x 10!/5!.

X

task 1: 5 seats in the NC segment can be occupied by 10 people
this can be done in 10x9x8x7x6=10!/5!(=30240) ways
task 2: 5 seats in the RC segment can be occupied by 10 people
this can be done in 10!/5! ways
task 3: 5 seats in the MC segment can be occupied by 10 people
this can be done in 10!/5! ways
now by applying multiplication principle of counting because actual task is achieved when all task 1 task 2 task 3 occur
so number of different ways they can sit= 10!/5! x 10!/5! x 10!/5!.

Permutation with restriction[Pending...] Hello everybody.
I have the following problem. Three irrigation events should implemented in a 19days...Ioannis 06 Dec 2018, 03:20

Independent vs non-mutually exclusive[Solved!] Section 9, example 2, it is stated that the events are not mutually exclusive.
Why are...phinah 03 Aug 2017, 04:30