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)
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 :-
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 \( 10656 + 3900 + 840 = 15396\) .
Re: Permutation of Flags
Posted: Fri Mar 28, 2014 11:35 pm
by Labib
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.
Re: Permutation of Flags
Posted: Sat Mar 29, 2014 7:14 pm
by sadman sakib
The line of confusion is :
Labib wrote:The next highest number of flags that belong to the same country is \(4\)
'same' always means more than \(1\) ,doesn't it ?
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
Posted: Sat Mar 29, 2014 9:47 pm
by Labib
Only one country has 4 flags.
I'll ask Saumitra da to check the solution.
Re: Permutation of Flags
Posted: Mon Apr 07, 2014 10:19 am
by Labib
Labib wrote:Only one country has 4 flags.
I'll ask Saumitra da to check the solution.
Saumitra da says, more than one country can have 4 flags.
sadman sakib wrote:
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 :-
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 \( 10656 + 3900 + 840 = 15396\) .
Saumitra da says, even according the given statement, the logics you used in some cases were wrong.
Try harder
