T20 World Cup 2014 is being held in Bangladesh.

Next to Mirpur stadium, there are $24$ flags standing in a row. The flags belong to $9$ different countries.

There are $10$ Australian flag. It's the highest number of flags a country has in that row.

The next highest number of flags that belong to the same country is $4$.

Pakistan and India have the same number of flags, but the number of flags each of them have is less than $4$.

How many permutations are possible for the flags?

(It is possible for a country to have $0$ flags)

Problem Setter: Saumitra da.

## Permutation of Flags

### Permutation of Flags

Please

**Install $L^AT_EX$ fonts**in your PC for better looking equations,**Learn****how to write equations**, and**don't forget**to read**Forum Guide and Rules.****"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes**-
**Posts:**17**Joined:**Thu Aug 29, 2013 6:33 pm**Location:**Japan Garden City, Mohammadpur, Dhaka

### Re: Permutation of Flags

There are \(10\) Australian flag . So , \(14\) flags belong to \(8\) different countries . Since , at least one country has \(4\) flags , there are three cases :-

So, \(1\) country among \(6\) countries (excluding Pakistan and India )can have these \(4\) flags in \(6\) ways . Then , the other \(10\) flags belong to \(7\) countries (including Pakistan and India ) . Now applying

\( \displaystyle \binom{5 + 10 - 1}{10} = 1001\)

\( \displaystyle \binom{5 + 8 - 1}{8} = 495\)

\( \displaystyle \binom{5 + 6 - 1}{6} = 210\)

\( \displaystyle \binom{5 + 4 - 1}{4} = 70\)

The number of permutation for this case is \( 6(1001 + 495 + 210 + 70) = 10656\) .

So, \(2\) countries among \(6\) countries (excluding Pakistan and India )can have these \(8\) flags in \( ^6P_2\) or \(30\) ways . Then , the other \(6\) flags belong to \(6\) countries (including Pakistan and India ) . Now applying

\( \displaystyle \binom{4 + 6 - 1}{6} = 84\)

\( \displaystyle \binom{4 + 4 - 1}{4} = 35\)

\( \displaystyle \binom{4 + 2 - 1}{2} = 10\)

The number of permutation for this case is \( 30(84 +35 + 10 + 1) = 3900\) .

So , \(3\) countries among \(6\) countries can have these \(12\) flags in \( ^6P_3\) or \(120\) ways . So , the other \(2\) flags belong to \(5\) countries . Again applying

\( \displaystyle \binom{3 + 2 - 1}{2} = 6\)

The number of permutation for this case would be \( 120(6 + 1) = 840\) .

Hence , the total number of possible permutation is \( 10656 + 3900 + 840 = 15396\) .

**Case 1 :**\(1\) country has \(4\) flags .So, \(1\) country among \(6\) countries (excluding Pakistan and India )can have these \(4\) flags in \(6\) ways . Then , the other \(10\) flags belong to \(7\) countries (including Pakistan and India ) . Now applying

*stars and bars*, we get the following cases :**Case 1.1 :**Pakistan = India = \(0\) . Then, \(5\) countries have \(10\) flags . The number of ways are :\( \displaystyle \binom{5 + 10 - 1}{10} = 1001\)

**Case 1.2 :**Pakistan = India = \(1\) . Then, \(5\) countries have \(8\) flags . The number of ways are :\( \displaystyle \binom{5 + 8 - 1}{8} = 495\)

**Case 1.3 :**Pakistan = India = \(2\) . Then, \(5\) countries have \(6\) flags . The number of ways are :\( \displaystyle \binom{5 + 6 - 1}{6} = 210\)

**Case 1.4 :**Pakistan = India = \(3\) . Then, \(5\) countries have \(4\) flags . The number of ways are :\( \displaystyle \binom{5 + 4 - 1}{4} = 70\)

The number of permutation for this case is \( 6(1001 + 495 + 210 + 70) = 10656\) .

**Case 2 :**\(2\) countries have \(4\) flags each .So, \(2\) countries among \(6\) countries (excluding Pakistan and India )can have these \(8\) flags in \( ^6P_2\) or \(30\) ways . Then , the other \(6\) flags belong to \(6\) countries (including Pakistan and India ) . Now applying

*stars and bars*, we get the following cases :**Case 2.1 :**Pakistan = India = \(0\) . Then, \(4\) countries have \(6\) flags . The number of ways are :\( \displaystyle \binom{4 + 6 - 1}{6} = 84\)

**Case 2.2 :**Pakistan = India = \(1\) . Then, \(4\) countries have \(4\) flags . The number of ways are :\( \displaystyle \binom{4 + 4 - 1}{4} = 35\)

**Case 2.3 :**Pakistan = India = \(2\) . Then, \(4\) countries have \(2\) flags . The number of ways are :\( \displaystyle \binom{4 + 2 - 1}{2} = 10\)

**Case 2.4 :**Pakistan = India = \(3\) . Here's just \(1\) permutation in this case .The number of permutation for this case is \( 30(84 +35 + 10 + 1) = 3900\) .

**Case 3 :**\(3\) countries have \(4\) flags each .So , \(3\) countries among \(6\) countries can have these \(12\) flags in \( ^6P_3\) or \(120\) ways . So , the other \(2\) flags belong to \(5\) countries . Again applying

*stars and bars*, we get the following cases :**Case 3.1 :**Pakistan = India = \(0\) . Then, \(3\) countries have \(2\) flags . The number of ways are :\( \displaystyle \binom{3 + 2 - 1}{2} = 6\)

**Case 3.2 :**Pakistan = India = \(1\) . Here's just \(1\) permutation in this case .The number of permutation for this case would be \( 120(6 + 1) = 840\) .

Hence , the total number of possible permutation is \( 10656 + 3900 + 840 = 15396\) .

Last edited by sadman sakib on Sat Mar 29, 2014 7:12 pm, edited 1 time in total.

### Re: Permutation of Flags

I think you got the problem wrong.

Australia has 10 flags. Another country has 4 flags. This country is neither India, nor Pakistan. It is the second highest.

The last 10 flags belong to the other 7 teams (which include India and Pakistan).

We are looking for the number of permutations possible.

I think the ans is a big number. You can change the number of second highest from 4 to 7.

Australia has 10 flags. Another country has 4 flags. This country is neither India, nor Pakistan. It is the second highest.

The last 10 flags belong to the other 7 teams (which include India and Pakistan).

We are looking for the number of permutations possible.

I think the ans is a big number. You can change the number of second highest from 4 to 7.

Please

**Install $L^AT_EX$ fonts**in your PC for better looking equations,**Learn****how to write equations**, and**don't forget**to read**Forum Guide and Rules.****"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes**-
**Posts:**17**Joined:**Thu Aug 29, 2013 6:33 pm**Location:**Japan Garden City, Mohammadpur, Dhaka

### Re: Permutation of Flags

The line of confusion is :

From your post it seems that you wanted to say : The next highest number of flags that belong to at least \(1\) country is \(4\) .Those two statements are quite different .

I just edited the answer accordingly . I think this is the right answer .

'same' always means more than \(1\) ,doesn't it ?Labib wrote:The next highest number of flags that belong to the same country is \(4\)

From your post it seems that you wanted to say : The next highest number of flags that belong to at least \(1\) country is \(4\) .Those two statements are quite different .

I just edited the answer accordingly . I think this is the right answer .

### Re: Permutation of Flags

Only one country has 4 flags.

I'll ask Saumitra da to check the solution.

I'll ask Saumitra da to check the solution.

Please

**Install $L^AT_EX$ fonts**in your PC for better looking equations,**Learn****how to write equations**, and**don't forget**to read**Forum Guide and Rules.****"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes**### Re: Permutation of Flags

Saumitra da says, more than one country can have 4 flags.Labib wrote:Only one country has 4 flags.

I'll ask Saumitra da to check the solution.

Saumitra da says, even according the given statement, the logics you used in some cases were wrong.sadman sakib wrote:

Try harder

**Install $L^AT_EX$ fonts**in your PC for better looking equations,

**Learn**

**how to write equations**, and

**don't forget**to read

**Forum Guide and Rules.**