# Combinatorics [Pending...]

**alex.vollenga** 19 May 2017, 09:58

### My question

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

3. Permutations

### 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!

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!

## Re: Combinatorics

**Murray** 24 May 2017, 01:28

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).

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'.

In each case, the remaining 10 need to be thought of as 6 "units" (the 5 from the same country who can sit anywhere in the remaining 10 chairs, plus 1 "unit" which is the block from the other country that can only sit in 5 of the reamining chairs.

I think I got all that right!

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'.
In each case, the remaining 10 need to be thought of as 6 "units" (the 5 from the same country who can sit anywhere in the remaining 10 chairs, plus 1 "unit" which is the block from the other country that can only sit in 5 of the reamining chairs.
I think I got all that right!
Regards
Murray

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