Mixing
How many ways can 26 letters be divided into 13 pairs?
$begingroup$
The question is: a plug board has 26 letters, and there are 13 cables. The cables connect all possible pair of letters.
How many possible configurations does the plug board provide? In other words, how many ways can 26 letters be divided into 13 pairs?
The result I found by calculation and some logic is:
$$25cdot 23cdot 21cdot 19cdot 17cdot 15cdot 13cdot 11cdot 9cdot 7cdot 5cdot 3 = 7 905 853 580 625.$$
Is this correct?
How is the mathematical algorithm and formula to calculate this and similar problems?
probability permutations combinations
edited Feb 1 at 17:43
Robert Z
101k1072145
asked Feb 1 at 17:18
Coder88Coder88
1066
$endgroup$
add a comment |
$begingroup$
The question is: a plug board has 26 letters, and there are 13 cables. The cables connect all possible pair of letters.
How many possible configurations does the plug board provide? In other words, how many ways can 26 letters be divided into 13 pairs?
The result I found by calculation and some logic is:
$$25cdot 23cdot 21cdot 19cdot 17cdot 15cdot 13cdot 11cdot 9cdot 7cdot 5cdot 3 = 7 905 853 580 625.$$
Is this correct?
How is the mathematical algorithm and formula to calculate this and similar problems?
probability permutations combinations
edited Feb 1 at 17:43
Robert Z
101k1072145
asked Feb 1 at 17:18
Coder88Coder88
1066
$endgroup$
$begingroup$
Your value can also be written as $frac {26!}{2^{13} cdot 13!}$
$endgroup$
– Mohammad Zuhair Khan
Feb 1 at 17:30
add a comment |
$begingroup$
The question is: a plug board has 26 letters, and there are 13 cables. The cables connect all possible pair of letters.
How many possible configurations does the plug board provide? In other words, how many ways can 26 letters be divided into 13 pairs?
The result I found by calculation and some logic is:
$$25cdot 23cdot 21cdot 19cdot 17cdot 15cdot 13cdot 11cdot 9cdot 7cdot 5cdot 3 = 7 905 853 580 625.$$
Is this correct?
How is the mathematical algorithm and formula to calculate this and similar problems?
probability permutations combinations
edited Feb 1 at 17:43
Robert Z
101k1072145
asked Feb 1 at 17:18
Coder88Coder88
1066
$endgroup$
The question is: a plug board has 26 letters, and there are 13 cables. The cables connect all possible pair of letters.
How many possible configurations does the plug board provide? In other words, how many ways can 26 letters be divided into 13 pairs?
The result I found by calculation and some logic is:
$$25cdot 23cdot 21cdot 19cdot 17cdot 15cdot 13cdot 11cdot 9cdot 7cdot 5cdot 3 = 7 905 853 580 625.$$
Is this correct?
How is the mathematical algorithm and formula to calculate this and similar problems?
probability permutations combinations
probability permutations combinations
edited Feb 1 at 17:43
Robert Z
101k1072145
asked Feb 1 at 17:18
Coder88Coder88
1066
edited Feb 1 at 17:43
Robert Z
101k1072145
asked Feb 1 at 17:18
Coder88Coder88
1066
edited Feb 1 at 17:43
Robert Z
101k1072145
edited Feb 1 at 17:43
Robert Z
101k1072145
edited Feb 1 at 17:43
Robert Z
101k1072145
101k1072145
asked Feb 1 at 17:18
Coder88Coder88
1066
asked Feb 1 at 17:18
Coder88Coder88
1066
asked Feb 1 at 17:18
Coder88Coder88
1066
1066
$begingroup$
Your value can also be written as $frac {26!}{2^{13} cdot 13!}$
$endgroup$
– Mohammad Zuhair Khan
Feb 1 at 17:30
add a comment |
$begingroup$
Your value can also be written as $frac {26!}{2^{13} cdot 13!}$
$endgroup$
– Mohammad Zuhair Khan
Feb 1 at 17:30
$begingroup$
Your value can also be written as $frac {26!}{2^{13} cdot 13!}$
$endgroup$
– Mohammad Zuhair Khan
Feb 1 at 17:30
$begingroup$
Your value can also be written as $frac {26!}{2^{13} cdot 13!}$
$endgroup$
– Mohammad Zuhair Khan
Feb 1 at 17:30
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Yes, you are correct. This is my approach: we select a set of $13$ letters among the $26$ in $binom{26}{13}$ ways. Then we assign the remaining $13$ letters to the chosen letters in $13!$ ways. Finally we divide the result by $2$, $13$ times because we are not interested in the order of the letters in the $13$ couples. The final result is
$$frac{binom{26}{13}cdot 13!}{2^{13}}=frac {26!}{2^{13} cdot 13!}=7 905 853 580 625$$
which is the same of yours.
answered Feb 1 at 17:31
Robert ZRobert Z
101k1072145
$endgroup$
$begingroup$
That was more simpler and smarter approach. My approach was daunting, I just counted by hand all possibilities with just 2 letters up to 6 letters using my fingers (all possible combinations of pairing 6 letters), found a logic and pattern, and used that to come with solution to the problem "26 letters and 13 cables".
$endgroup$
– Coder88
Feb 1 at 21:22
add a comment |
$begingroup$
One way to count would be as follows.
Consider the letters ordered, and labeled as $1,2,3,4,dots,2N$.
Let $A(N)$ be the number of possibilities to "pair them in $N$ pairs".
We take the first letter, $1$, and pair it, i.e. we count its possible pairings.
There are $(2N-1)$ possibilities. We make a choice of the $1$-match, eliminate this pair from the list, reorder, and count again.
In this way we get inductively:
$$
A(2N)=(2N-1)A(2N-2) .
$$
Because $A(2)=1$, we get the formula $A(2N)=(2N-1)(2N-3)cdots5cdot 3cdot 1$.
(Some people call this number $(2N-1)!!$.)
(It is hard to put the hands on the "similar problems" part of the question.)
answered Feb 1 at 17:31
dan_fuleadan_fulea
7,1781513
$endgroup$
add a comment |
$begingroup$
Another approach is: the first pair can be selected in ${26choose 2}$ ways, but we are not interested in the position of the pair among the $13$ pairs, hence we divide it by ${13choose 1}$:
$$frac{{26choose 2}}{{13choose 1}}$$
Similarly, for other pairs.
At the end we get:
$$frac{{26choose 2}}{{13choose 1}}cdot frac{{24choose 2}}{{12choose 1}}cdotfrac{{22choose 2}}{{11choose 1}}cdotfrac{{20choose 2}}{{10choose 1}}cdot frac{{18choose 2}}{{9choose 1}}cdotfrac{{16choose 2}}{{8choose 1}}cdotfrac{{14choose 2}}{{7choose 1}}cdotfrac{{12choose 2}}{{6choose 1}}cdotfrac{{10choose 2}}{{5choose 1}}cdotfrac{{8choose 2}}{{4choose 1}}cdotfrac{{6choose 2}}{{3choose 1}}cdotfrac{{4choose 2}}{{2choose 1}}cdotfrac{{2choose 2}}{{1choose 1}},$$
which is what you have.
answered Feb 1 at 18:45
farruhotafarruhota
21.9k2942
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, you are correct. This is my approach: we select a set of $13$ letters among the $26$ in $binom{26}{13}$ ways. Then we assign the remaining $13$ letters to the chosen letters in $13!$ ways. Finally we divide the result by $2$, $13$ times because we are not interested in the order of the letters in the $13$ couples. The final result is
$$frac{binom{26}{13}cdot 13!}{2^{13}}=frac {26!}{2^{13} cdot 13!}=7 905 853 580 625$$
which is the same of yours.
answered Feb 1 at 17:31
Robert ZRobert Z
101k1072145
$endgroup$
$begingroup$
That was more simpler and smarter approach. My approach was daunting, I just counted by hand all possibilities with just 2 letters up to 6 letters using my fingers (all possible combinations of pairing 6 letters), found a logic and pattern, and used that to come with solution to the problem "26 letters and 13 cables".
$endgroup$
– Coder88
Feb 1 at 21:22
add a comment |
$begingroup$
Yes, you are correct. This is my approach: we select a set of $13$ letters among the $26$ in $binom{26}{13}$ ways. Then we assign the remaining $13$ letters to the chosen letters in $13!$ ways. Finally we divide the result by $2$, $13$ times because we are not interested in the order of the letters in the $13$ couples. The final result is
$$frac{binom{26}{13}cdot 13!}{2^{13}}=frac {26!}{2^{13} cdot 13!}=7 905 853 580 625$$
which is the same of yours.
answered Feb 1 at 17:31
Robert ZRobert Z
101k1072145
$endgroup$
$begingroup$
That was more simpler and smarter approach. My approach was daunting, I just counted by hand all possibilities with just 2 letters up to 6 letters using my fingers (all possible combinations of pairing 6 letters), found a logic and pattern, and used that to come with solution to the problem "26 letters and 13 cables".
$endgroup$
– Coder88
Feb 1 at 21:22
add a comment |
$begingroup$
Yes, you are correct. This is my approach: we select a set of $13$ letters among the $26$ in $binom{26}{13}$ ways. Then we assign the remaining $13$ letters to the chosen letters in $13!$ ways. Finally we divide the result by $2$, $13$ times because we are not interested in the order of the letters in the $13$ couples. The final result is
$$frac{binom{26}{13}cdot 13!}{2^{13}}=frac {26!}{2^{13} cdot 13!}=7 905 853 580 625$$
which is the same of yours.
answered Feb 1 at 17:31
Robert ZRobert Z
101k1072145
$endgroup$
Yes, you are correct. This is my approach: we select a set of $13$ letters among the $26$ in $binom{26}{13}$ ways. Then we assign the remaining $13$ letters to the chosen letters in $13!$ ways. Finally we divide the result by $2$, $13$ times because we are not interested in the order of the letters in the $13$ couples. The final result is
$$frac{binom{26}{13}cdot 13!}{2^{13}}=frac {26!}{2^{13} cdot 13!}=7 905 853 580 625$$
which is the same of yours.
answered Feb 1 at 17:31
Robert ZRobert Z
101k1072145
edited Feb 1 at 17:43
answered Feb 1 at 17:31
Robert ZRobert Z
101k1072145
answered Feb 1 at 17:31
Robert ZRobert Z
101k1072145
answered Feb 1 at 17:31
Robert ZRobert Z
101k1072145
101k1072145
$begingroup$
That was more simpler and smarter approach. My approach was daunting, I just counted by hand all possibilities with just 2 letters up to 6 letters using my fingers (all possible combinations of pairing 6 letters), found a logic and pattern, and used that to come with solution to the problem "26 letters and 13 cables".
$endgroup$
– Coder88
Feb 1 at 21:22
add a comment |
$begingroup$
That was more simpler and smarter approach. My approach was daunting, I just counted by hand all possibilities with just 2 letters up to 6 letters using my fingers (all possible combinations of pairing 6 letters), found a logic and pattern, and used that to come with solution to the problem "26 letters and 13 cables".
$endgroup$
– Coder88
Feb 1 at 21:22
$begingroup$
That was more simpler and smarter approach. My approach was daunting, I just counted by hand all possibilities with just 2 letters up to 6 letters using my fingers (all possible combinations of pairing 6 letters), found a logic and pattern, and used that to come with solution to the problem "26 letters and 13 cables".
$endgroup$
– Coder88
Feb 1 at 21:22
$begingroup$
That was more simpler and smarter approach. My approach was daunting, I just counted by hand all possibilities with just 2 letters up to 6 letters using my fingers (all possible combinations of pairing 6 letters), found a logic and pattern, and used that to come with solution to the problem "26 letters and 13 cables".
$endgroup$
– Coder88
Feb 1 at 21:22
add a comment |
$begingroup$
One way to count would be as follows.
Consider the letters ordered, and labeled as $1,2,3,4,dots,2N$.
Let $A(N)$ be the number of possibilities to "pair them in $N$ pairs".
We take the first letter, $1$, and pair it, i.e. we count its possible pairings.
There are $(2N-1)$ possibilities. We make a choice of the $1$-match, eliminate this pair from the list, reorder, and count again.
In this way we get inductively:
$$
A(2N)=(2N-1)A(2N-2) .
$$
Because $A(2)=1$, we get the formula $A(2N)=(2N-1)(2N-3)cdots5cdot 3cdot 1$.
(Some people call this number $(2N-1)!!$.)
(It is hard to put the hands on the "similar problems" part of the question.)
answered Feb 1 at 17:31
dan_fuleadan_fulea
7,1781513
$endgroup$
add a comment |
$begingroup$
One way to count would be as follows.
Consider the letters ordered, and labeled as $1,2,3,4,dots,2N$.
Let $A(N)$ be the number of possibilities to "pair them in $N$ pairs".
We take the first letter, $1$, and pair it, i.e. we count its possible pairings.
There are $(2N-1)$ possibilities. We make a choice of the $1$-match, eliminate this pair from the list, reorder, and count again.
In this way we get inductively:
$$
A(2N)=(2N-1)A(2N-2) .
$$
Because $A(2)=1$, we get the formula $A(2N)=(2N-1)(2N-3)cdots5cdot 3cdot 1$.
(Some people call this number $(2N-1)!!$.)
(It is hard to put the hands on the "similar problems" part of the question.)
answered Feb 1 at 17:31
dan_fuleadan_fulea
7,1781513
$endgroup$
add a comment |
$begingroup$
One way to count would be as follows.
Consider the letters ordered, and labeled as $1,2,3,4,dots,2N$.
Let $A(N)$ be the number of possibilities to "pair them in $N$ pairs".
We take the first letter, $1$, and pair it, i.e. we count its possible pairings.
There are $(2N-1)$ possibilities. We make a choice of the $1$-match, eliminate this pair from the list, reorder, and count again.
In this way we get inductively:
$$
A(2N)=(2N-1)A(2N-2) .
$$
Because $A(2)=1$, we get the formula $A(2N)=(2N-1)(2N-3)cdots5cdot 3cdot 1$.
(Some people call this number $(2N-1)!!$.)
(It is hard to put the hands on the "similar problems" part of the question.)
answered Feb 1 at 17:31
dan_fuleadan_fulea
7,1781513
$endgroup$
One way to count would be as follows.
Consider the letters ordered, and labeled as $1,2,3,4,dots,2N$.
Let $A(N)$ be the number of possibilities to "pair them in $N$ pairs".
We take the first letter, $1$, and pair it, i.e. we count its possible pairings.
There are $(2N-1)$ possibilities. We make a choice of the $1$-match, eliminate this pair from the list, reorder, and count again.
In this way we get inductively:
$$
A(2N)=(2N-1)A(2N-2) .
$$
Because $A(2)=1$, we get the formula $A(2N)=(2N-1)(2N-3)cdots5cdot 3cdot 1$.
(Some people call this number $(2N-1)!!$.)
(It is hard to put the hands on the "similar problems" part of the question.)
answered Feb 1 at 17:31
dan_fuleadan_fulea
7,1781513
answered Feb 1 at 17:31
dan_fuleadan_fulea
7,1781513
answered Feb 1 at 17:31
dan_fuleadan_fulea
7,1781513
answered Feb 1 at 17:31
dan_fuleadan_fulea
7,1781513
7,1781513
add a comment |
add a comment |
$begingroup$
Another approach is: the first pair can be selected in ${26choose 2}$ ways, but we are not interested in the position of the pair among the $13$ pairs, hence we divide it by ${13choose 1}$:
$$frac{{26choose 2}}{{13choose 1}}$$
Similarly, for other pairs.
At the end we get:
$$frac{{26choose 2}}{{13choose 1}}cdot frac{{24choose 2}}{{12choose 1}}cdotfrac{{22choose 2}}{{11choose 1}}cdotfrac{{20choose 2}}{{10choose 1}}cdot frac{{18choose 2}}{{9choose 1}}cdotfrac{{16choose 2}}{{8choose 1}}cdotfrac{{14choose 2}}{{7choose 1}}cdotfrac{{12choose 2}}{{6choose 1}}cdotfrac{{10choose 2}}{{5choose 1}}cdotfrac{{8choose 2}}{{4choose 1}}cdotfrac{{6choose 2}}{{3choose 1}}cdotfrac{{4choose 2}}{{2choose 1}}cdotfrac{{2choose 2}}{{1choose 1}},$$
which is what you have.
answered Feb 1 at 18:45
farruhotafarruhota
21.9k2942
$endgroup$
add a comment |
$begingroup$
Another approach is: the first pair can be selected in ${26choose 2}$ ways, but we are not interested in the position of the pair among the $13$ pairs, hence we divide it by ${13choose 1}$:
$$frac{{26choose 2}}{{13choose 1}}$$
Similarly, for other pairs.
At the end we get:
$$frac{{26choose 2}}{{13choose 1}}cdot frac{{24choose 2}}{{12choose 1}}cdotfrac{{22choose 2}}{{11choose 1}}cdotfrac{{20choose 2}}{{10choose 1}}cdot frac{{18choose 2}}{{9choose 1}}cdotfrac{{16choose 2}}{{8choose 1}}cdotfrac{{14choose 2}}{{7choose 1}}cdotfrac{{12choose 2}}{{6choose 1}}cdotfrac{{10choose 2}}{{5choose 1}}cdotfrac{{8choose 2}}{{4choose 1}}cdotfrac{{6choose 2}}{{3choose 1}}cdotfrac{{4choose 2}}{{2choose 1}}cdotfrac{{2choose 2}}{{1choose 1}},$$
which is what you have.
answered Feb 1 at 18:45
farruhotafarruhota
21.9k2942
$endgroup$
add a comment |
$begingroup$
Another approach is: the first pair can be selected in ${26choose 2}$ ways, but we are not interested in the position of the pair among the $13$ pairs, hence we divide it by ${13choose 1}$:
$$frac{{26choose 2}}{{13choose 1}}$$
Similarly, for other pairs.
At the end we get:
$$frac{{26choose 2}}{{13choose 1}}cdot frac{{24choose 2}}{{12choose 1}}cdotfrac{{22choose 2}}{{11choose 1}}cdotfrac{{20choose 2}}{{10choose 1}}cdot frac{{18choose 2}}{{9choose 1}}cdotfrac{{16choose 2}}{{8choose 1}}cdotfrac{{14choose 2}}{{7choose 1}}cdotfrac{{12choose 2}}{{6choose 1}}cdotfrac{{10choose 2}}{{5choose 1}}cdotfrac{{8choose 2}}{{4choose 1}}cdotfrac{{6choose 2}}{{3choose 1}}cdotfrac{{4choose 2}}{{2choose 1}}cdotfrac{{2choose 2}}{{1choose 1}},$$
which is what you have.
answered Feb 1 at 18:45
farruhotafarruhota
21.9k2942
$endgroup$
Another approach is: the first pair can be selected in ${26choose 2}$ ways, but we are not interested in the position of the pair among the $13$ pairs, hence we divide it by ${13choose 1}$:
$$frac{{26choose 2}}{{13choose 1}}$$
Similarly, for other pairs.
At the end we get:
$$frac{{26choose 2}}{{13choose 1}}cdot frac{{24choose 2}}{{12choose 1}}cdotfrac{{22choose 2}}{{11choose 1}}cdotfrac{{20choose 2}}{{10choose 1}}cdot frac{{18choose 2}}{{9choose 1}}cdotfrac{{16choose 2}}{{8choose 1}}cdotfrac{{14choose 2}}{{7choose 1}}cdotfrac{{12choose 2}}{{6choose 1}}cdotfrac{{10choose 2}}{{5choose 1}}cdotfrac{{8choose 2}}{{4choose 1}}cdotfrac{{6choose 2}}{{3choose 1}}cdotfrac{{4choose 2}}{{2choose 1}}cdotfrac{{2choose 2}}{{1choose 1}},$$
which is what you have.
answered Feb 1 at 18:45
farruhotafarruhota
21.9k2942
answered Feb 1 at 18:45
farruhotafarruhota
21.9k2942
answered Feb 1 at 18:45
farruhotafarruhota
21.9k2942
answered Feb 1 at 18:45
farruhotafarruhota
21.9k2942
21.9k2942
add a comment |
add a comment |
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$begingroup$
Your value can also be written as $frac {26!}{2^{13} cdot 13!}$
$endgroup$
– Mohammad Zuhair Khan
Feb 1 at 17:30