Mixing
How to prove $binom{n}{4}=sum_{i=3}^{n-1} binom{i}{3}$?
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I was trying to find the number of diagonals of a nth polygon by deduction. Let $a_n$ be the number of the diagonals of thr nth polygon. Then I've got $$a_{n+1}=a_n+binom{n}{3}$$
$$a_n=sum_{i=3}^{n-1} binom{i}{3}$$
But we simply know that $a_n=binom{n}{4}$, where $a_n$ is the number of diagonals.
So we get $binom{n}{4}=sum_{i=3}^{n-2} binom{i}{3}$. My question is whether it is a trivial equation or it can be generalized like $binom{n}{k}=sum_{i=k-1}^{n-1} binom{i}{k-1}$. If it can be, how to prove that?
combinatorics
asked Jan 15 at 14:00
Mahmud Nabil SnigdhoMahmud Nabil Snigdho
12
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add a comment |
$begingroup$
I was trying to find the number of diagonals of a nth polygon by deduction. Let $a_n$ be the number of the diagonals of thr nth polygon. Then I've got $$a_{n+1}=a_n+binom{n}{3}$$
$$a_n=sum_{i=3}^{n-1} binom{i}{3}$$
But we simply know that $a_n=binom{n}{4}$, where $a_n$ is the number of diagonals.
So we get $binom{n}{4}=sum_{i=3}^{n-2} binom{i}{3}$. My question is whether it is a trivial equation or it can be generalized like $binom{n}{k}=sum_{i=k-1}^{n-1} binom{i}{k-1}$. If it can be, how to prove that?
combinatorics
asked Jan 15 at 14:00
Mahmud Nabil SnigdhoMahmud Nabil Snigdho
12
$endgroup$
$begingroup$
$$binom nk-binom{n-1}k=?$$ math.stackexchange.com/questions/1125923/…
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– lab bhattacharjee
Jan 15 at 14:03
4
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This is the Hockey-stick_identity: see en.wikipedia.org/wiki/Hockey-stick_identity
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– Robert Z
Jan 15 at 14:03
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Can be proved by induction.
$endgroup$
– gandalf61
Jan 15 at 14:11
add a comment |
$begingroup$
I was trying to find the number of diagonals of a nth polygon by deduction. Let $a_n$ be the number of the diagonals of thr nth polygon. Then I've got $$a_{n+1}=a_n+binom{n}{3}$$
$$a_n=sum_{i=3}^{n-1} binom{i}{3}$$
But we simply know that $a_n=binom{n}{4}$, where $a_n$ is the number of diagonals.
So we get $binom{n}{4}=sum_{i=3}^{n-2} binom{i}{3}$. My question is whether it is a trivial equation or it can be generalized like $binom{n}{k}=sum_{i=k-1}^{n-1} binom{i}{k-1}$. If it can be, how to prove that?
combinatorics
asked Jan 15 at 14:00
Mahmud Nabil SnigdhoMahmud Nabil Snigdho
12
$endgroup$
I was trying to find the number of diagonals of a nth polygon by deduction. Let $a_n$ be the number of the diagonals of thr nth polygon. Then I've got $$a_{n+1}=a_n+binom{n}{3}$$
$$a_n=sum_{i=3}^{n-1} binom{i}{3}$$
But we simply know that $a_n=binom{n}{4}$, where $a_n$ is the number of diagonals.
So we get $binom{n}{4}=sum_{i=3}^{n-2} binom{i}{3}$. My question is whether it is a trivial equation or it can be generalized like $binom{n}{k}=sum_{i=k-1}^{n-1} binom{i}{k-1}$. If it can be, how to prove that?
combinatorics
combinatorics
asked Jan 15 at 14:00
Mahmud Nabil SnigdhoMahmud Nabil Snigdho
12
asked Jan 15 at 14:00
Mahmud Nabil SnigdhoMahmud Nabil Snigdho
12
asked Jan 15 at 14:00
Mahmud Nabil SnigdhoMahmud Nabil Snigdho
12
asked Jan 15 at 14:00
Mahmud Nabil SnigdhoMahmud Nabil Snigdho
12
asked Jan 15 at 14:00
Mahmud Nabil SnigdhoMahmud Nabil Snigdho
12
12
$begingroup$
$$binom nk-binom{n-1}k=?$$ math.stackexchange.com/questions/1125923/…
$endgroup$
– lab bhattacharjee
Jan 15 at 14:03
4
$begingroup$
This is the Hockey-stick_identity: see en.wikipedia.org/wiki/Hockey-stick_identity
$endgroup$
– Robert Z
Jan 15 at 14:03
$begingroup$
Can be proved by induction.
$endgroup$
– gandalf61
Jan 15 at 14:11
add a comment |
$begingroup$
$$binom nk-binom{n-1}k=?$$ math.stackexchange.com/questions/1125923/…
$endgroup$
– lab bhattacharjee
Jan 15 at 14:03
4
$begingroup$
This is the Hockey-stick_identity: see en.wikipedia.org/wiki/Hockey-stick_identity
$endgroup$
– Robert Z
Jan 15 at 14:03
$begingroup$
Can be proved by induction.
$endgroup$
– gandalf61
Jan 15 at 14:11
$begingroup$
$$binom nk-binom{n-1}k=?$$ math.stackexchange.com/questions/1125923/…
$endgroup$
– lab bhattacharjee
Jan 15 at 14:03
$begingroup$
$$binom nk-binom{n-1}k=?$$ math.stackexchange.com/questions/1125923/…
$endgroup$
– lab bhattacharjee
Jan 15 at 14:03
4
4
$begingroup$
This is the Hockey-stick_identity: see en.wikipedia.org/wiki/Hockey-stick_identity
$endgroup$
– Robert Z
Jan 15 at 14:03
$begingroup$
This is the Hockey-stick_identity: see en.wikipedia.org/wiki/Hockey-stick_identity
$endgroup$
– Robert Z
Jan 15 at 14:03
$begingroup$
Can be proved by induction.
$endgroup$
– gandalf61
Jan 15 at 14:11
$begingroup$
Can be proved by induction.
$endgroup$
– gandalf61
Jan 15 at 14:11
add a comment |
3 Answers
3
active
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$begingroup$
You may transform the sum into a telescoping sum using the "binomial" fact
- $binom{i}{3} = binom{i+1}{4} - binom{i}{4}$
Hence,
begin{eqnarray*} sum_{i=3}^{n-1} binom{i}{3}
& = & sum_{i=3}^{n-1} left(binom{i+1}{4} - binom{i}{4} right)\
& stackrel{binom{3}{4} = 0}{=} & binom{n-1+1}{4} - 0\
& = & binom{n}{4}\
end{eqnarray*}
answered Jan 15 at 14:15
trancelocationtrancelocation
11.9k1825
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add a comment |
$begingroup$
We have to choose four numbers from $[n]$. Assume the largest chosen number is $i+1in[4..n]$. The other three numbers then are arbitrary three numbers in $[i]$.
answered Jan 15 at 14:33
Christian BlatterChristian Blatter
174k8115327
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add a comment |
$begingroup$
Let $k=3$ so that $4=k+1$. The binomial coefficient $binom n{k+1}$ counts the $k+1$-element subsets $S$ of the $n$-element set ${0,1,ldots,n-1}$. Let $S$ be such a set and $i=max(S)$. Then $kleq i<n$, and $S=S'cup{i}$ where $S'$ can be any $k$-element subset of the $i$-element set ${0,1,ldots,i-1}$. So
$$binom n{k+1}=sum_{i=k}^{n-1}binom ik.$$
answered Jan 15 at 14:34
Marc van LeeuwenMarc van Leeuwen
87.4k5109224
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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
$begingroup$
You may transform the sum into a telescoping sum using the "binomial" fact
- $binom{i}{3} = binom{i+1}{4} - binom{i}{4}$
Hence,
begin{eqnarray*} sum_{i=3}^{n-1} binom{i}{3}
& = & sum_{i=3}^{n-1} left(binom{i+1}{4} - binom{i}{4} right)\
& stackrel{binom{3}{4} = 0}{=} & binom{n-1+1}{4} - 0\
& = & binom{n}{4}\
end{eqnarray*}
answered Jan 15 at 14:15
trancelocationtrancelocation
11.9k1825
$endgroup$
add a comment |
$begingroup$
You may transform the sum into a telescoping sum using the "binomial" fact
- $binom{i}{3} = binom{i+1}{4} - binom{i}{4}$
Hence,
begin{eqnarray*} sum_{i=3}^{n-1} binom{i}{3}
& = & sum_{i=3}^{n-1} left(binom{i+1}{4} - binom{i}{4} right)\
& stackrel{binom{3}{4} = 0}{=} & binom{n-1+1}{4} - 0\
& = & binom{n}{4}\
end{eqnarray*}
answered Jan 15 at 14:15
trancelocationtrancelocation
11.9k1825
$endgroup$
add a comment |
$begingroup$
You may transform the sum into a telescoping sum using the "binomial" fact
- $binom{i}{3} = binom{i+1}{4} - binom{i}{4}$
Hence,
begin{eqnarray*} sum_{i=3}^{n-1} binom{i}{3}
& = & sum_{i=3}^{n-1} left(binom{i+1}{4} - binom{i}{4} right)\
& stackrel{binom{3}{4} = 0}{=} & binom{n-1+1}{4} - 0\
& = & binom{n}{4}\
end{eqnarray*}
answered Jan 15 at 14:15
trancelocationtrancelocation
11.9k1825
$endgroup$
You may transform the sum into a telescoping sum using the "binomial" fact
- $binom{i}{3} = binom{i+1}{4} - binom{i}{4}$
Hence,
begin{eqnarray*} sum_{i=3}^{n-1} binom{i}{3}
& = & sum_{i=3}^{n-1} left(binom{i+1}{4} - binom{i}{4} right)\
& stackrel{binom{3}{4} = 0}{=} & binom{n-1+1}{4} - 0\
& = & binom{n}{4}\
end{eqnarray*}
answered Jan 15 at 14:15
trancelocationtrancelocation
11.9k1825
answered Jan 15 at 14:15
trancelocationtrancelocation
11.9k1825
answered Jan 15 at 14:15
trancelocationtrancelocation
11.9k1825
answered Jan 15 at 14:15
trancelocationtrancelocation
11.9k1825
11.9k1825
add a comment |
add a comment |
$begingroup$
We have to choose four numbers from $[n]$. Assume the largest chosen number is $i+1in[4..n]$. The other three numbers then are arbitrary three numbers in $[i]$.
answered Jan 15 at 14:33
Christian BlatterChristian Blatter
174k8115327
$endgroup$
add a comment |
$begingroup$
We have to choose four numbers from $[n]$. Assume the largest chosen number is $i+1in[4..n]$. The other three numbers then are arbitrary three numbers in $[i]$.
answered Jan 15 at 14:33
Christian BlatterChristian Blatter
174k8115327
$endgroup$
add a comment |
$begingroup$
We have to choose four numbers from $[n]$. Assume the largest chosen number is $i+1in[4..n]$. The other three numbers then are arbitrary three numbers in $[i]$.
answered Jan 15 at 14:33
Christian BlatterChristian Blatter
174k8115327
$endgroup$
We have to choose four numbers from $[n]$. Assume the largest chosen number is $i+1in[4..n]$. The other three numbers then are arbitrary three numbers in $[i]$.
answered Jan 15 at 14:33
Christian BlatterChristian Blatter
174k8115327
answered Jan 15 at 14:33
Christian BlatterChristian Blatter
174k8115327
answered Jan 15 at 14:33
Christian BlatterChristian Blatter
174k8115327
answered Jan 15 at 14:33
Christian BlatterChristian Blatter
174k8115327
174k8115327
add a comment |
add a comment |
$begingroup$
Let $k=3$ so that $4=k+1$. The binomial coefficient $binom n{k+1}$ counts the $k+1$-element subsets $S$ of the $n$-element set ${0,1,ldots,n-1}$. Let $S$ be such a set and $i=max(S)$. Then $kleq i<n$, and $S=S'cup{i}$ where $S'$ can be any $k$-element subset of the $i$-element set ${0,1,ldots,i-1}$. So
$$binom n{k+1}=sum_{i=k}^{n-1}binom ik.$$
answered Jan 15 at 14:34
Marc van LeeuwenMarc van Leeuwen
87.4k5109224
$endgroup$
add a comment |
$begingroup$
Let $k=3$ so that $4=k+1$. The binomial coefficient $binom n{k+1}$ counts the $k+1$-element subsets $S$ of the $n$-element set ${0,1,ldots,n-1}$. Let $S$ be such a set and $i=max(S)$. Then $kleq i<n$, and $S=S'cup{i}$ where $S'$ can be any $k$-element subset of the $i$-element set ${0,1,ldots,i-1}$. So
$$binom n{k+1}=sum_{i=k}^{n-1}binom ik.$$
answered Jan 15 at 14:34
Marc van LeeuwenMarc van Leeuwen
87.4k5109224
$endgroup$
add a comment |
$begingroup$
Let $k=3$ so that $4=k+1$. The binomial coefficient $binom n{k+1}$ counts the $k+1$-element subsets $S$ of the $n$-element set ${0,1,ldots,n-1}$. Let $S$ be such a set and $i=max(S)$. Then $kleq i<n$, and $S=S'cup{i}$ where $S'$ can be any $k$-element subset of the $i$-element set ${0,1,ldots,i-1}$. So
$$binom n{k+1}=sum_{i=k}^{n-1}binom ik.$$
answered Jan 15 at 14:34
Marc van LeeuwenMarc van Leeuwen
87.4k5109224
$endgroup$
Let $k=3$ so that $4=k+1$. The binomial coefficient $binom n{k+1}$ counts the $k+1$-element subsets $S$ of the $n$-element set ${0,1,ldots,n-1}$. Let $S$ be such a set and $i=max(S)$. Then $kleq i<n$, and $S=S'cup{i}$ where $S'$ can be any $k$-element subset of the $i$-element set ${0,1,ldots,i-1}$. So
$$binom n{k+1}=sum_{i=k}^{n-1}binom ik.$$
answered Jan 15 at 14:34
Marc van LeeuwenMarc van Leeuwen
87.4k5109224
answered Jan 15 at 14:34
Marc van LeeuwenMarc van Leeuwen
87.4k5109224
answered Jan 15 at 14:34
Marc van LeeuwenMarc van Leeuwen
87.4k5109224
answered Jan 15 at 14:34
Marc van LeeuwenMarc van Leeuwen
87.4k5109224
87.4k5109224
add a comment |
add a comment |
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$begingroup$
$$binom nk-binom{n-1}k=?$$ math.stackexchange.com/questions/1125923/…
$endgroup$
– lab bhattacharjee
Jan 15 at 14:03
4
$begingroup$
This is the Hockey-stick_identity: see en.wikipedia.org/wiki/Hockey-stick_identity
$endgroup$
– Robert Z
Jan 15 at 14:03
$begingroup$
Can be proved by induction.
$endgroup$
– gandalf61
Jan 15 at 14:11