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# Combinatorics [Solved!]

### 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?

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

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'.

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

## Re: Combinatorics

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?

## Re: Combinatorics

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
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
so number of different ways they can sit= 10!/5! x 10!/5! x 10!/5!.

## Re: Combinatorics

I do not think that OP should have applied the third theorem from 3. Permutations in this Case. In this problem, each individual is unique so that theorem is does not apply.

I could not understand what Murray meant by, "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 followed stephenB but could not turn that into an answer.

AkashRajput is wrong. 10!/5! x 10!/5! x 10!/5! is greater than 15!. 15! would be the number of ways 15 people could fill 15 seats without any restrictions. The answer here must be less than 15!.

For my answer I will use the segments and NC NR NM notation that Murray suggested.

I break the problem up into 6 Cases:
5 russians, 0 mexicans in the NC seats (Case 1)
4 russians, 1 mexicans in the NC seats (Case 2)
3 russians, 2 mexicans in the NC seats (Case 3)
2 russians, 3 mexicans in the NC seats (Case 4)
1 russians, 4 mexicans in the NC seats (Case 5)
0 russians, 5 mexicans in the NC seats (Case 6)

In each Case, I start by filling the NC seats, then the NR seats, then the NM seats.

Case 1: 5 russians, 0 mexicans in the NC seats:
5! ways to seat the russians in the NC seats
5! ways to seat the mexicans in the NR seats
5! ways to seat the chinese in the NM seats
=> (5!)^3 arrangements in Case 1.

Case 2: 4 russians, 1 mexicans in the NC seats:
5*5*5! ways to fill the NC seats
5*5! ways to fill the NR seats
5! ways to fill the NM seats
=> (5!)^3 * 5^3 arrangements in Case 2.

Case 3: 3 russians, 2 mexicans in the NC seats:
5*4*5*4*5! ways to fill the NC seats
5*4*5! ways to fill the NR seats
5! ways to fill the NM seats
=> (5!)^3 * 5^3 * 4^3 arrangements in Case 3.

Case 1 and Case 6 have the same number of arrangements.
Case 2 and Case 5 have the same number of arrangements.
Case 3 and Case 4 have the same number of arrangements.
So we can just multiply what we've come up with by 2.

So my answer is 2 * (5!)^3 * (1 + 5^3 + 5^3*4^3).

Sanity check: 2 * (5!)^3 * (1 + 5^3 + 5^3*4^3) / 15! is roughly 0.02. That seems reasonable to me.

X

Hello, a few comments first:

I do not think that OP should have applied the third theorem from <a href="/counting-probability/3-permutations.php">3. Permutations</a> in this Case. In this problem, each individual is unique so that theorem is does not apply.

I could not understand what Murray meant by, <i>"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>

I followed stephenB but could not turn that into an answer.

AkashRajput is wrong. 10!/5! x 10!/5! x 10!/5! is greater than 15!. 15! would be the number of ways 15 people could fill 15 seats without any restrictions. The answer here must be less than 15!.

For my answer I will use the segments and NC NR NM notation that Murray suggested.

I break the problem up into 6 Cases:
5 russians, 0 mexicans in the NC seats (Case 1)
4 russians, 1 mexicans in the NC seats (Case 2)
3 russians, 2 mexicans in the NC seats (Case 3)
2 russians, 3 mexicans in the NC seats (Case 4)
1 russians, 4 mexicans in the NC seats (Case 5)
0 russians, 5 mexicans in the NC seats (Case 6)

In each Case, I start by filling the NC seats, then the NR seats, then the NM seats.

Case 1: 5 russians, 0 mexicans in the NC seats:
5! ways to seat the russians in the NC seats
5! ways to seat the mexicans in the NR seats
5! ways to seat the chinese in the NM seats
=&gt; (5!)^3 arrangements in Case 1.

Case 2: 4 russians, 1 mexicans in the NC seats:
5*5*5! ways to fill the NC seats
5*5! ways to fill the NR seats
5! ways to fill the NM seats
=&gt; (5!)^3 * 5^3 arrangements in Case 2.

Case 3: 3 russians, 2 mexicans in the NC seats:
5*4*5*4*5! ways to fill the NC seats
5*4*5! ways to fill the NR seats
5! ways to fill the NM seats
=&gt; (5!)^3 * 5^3 * 4^3 arrangements in Case 3.

Case 1 and Case 6 have the same number of arrangements.
Case 2 and Case 5 have the same number of arrangements.
Case 3 and Case 4 have the same number of arrangements.
So we can just multiply what we've come up with by 2.

So my answer is 2 * (5!)^3 * (1 + 5^3 + 5^3*4^3).

Sanity check: 2 * (5!)^3 * (1 + 5^3 + 5^3*4^3) / 15! is roughly 0.02. That seems reasonable to me.