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Difference between revisions of "Combinatorial identity"

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But we already know they can be picked in <math>\binom{n}{k}</math> ways, so  
 
But we already know they can be picked in <math>\binom{n}{k}</math> ways, so  
  
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<cmath>{n \choose k}={n-1\choose k-1}+{n-1\choose k} \qquad \qquad \square</cmath>
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<cmath>{n \choose k}={n-1\choose k-1}+{n-1\choose k} </cmath>
  
  
 
Also, we can look at Pascal's Triangle to see why this is. If we were to extend Pascal's Triangle to row n, we would see the term <math>\binom{n}{k}</math>. Above that, we would see the terms <math>{n-1\choose k-1}</math> and <math>{n-1\choose k}</math>. Due to the definition of Pascal's Triangle, <math>{n \choose k}={n-1\choose k-1}+{n-1\choose k}</math>.
 
Also, we can look at Pascal's Triangle to see why this is. If we were to extend Pascal's Triangle to row n, we would see the term <math>\binom{n}{k}</math>. Above that, we would see the terms <math>{n-1\choose k-1}</math> and <math>{n-1\choose k}</math>. Due to the definition of Pascal's Triangle, <math>{n \choose k}={n-1\choose k-1}+{n-1\choose k}</math>.
−
 
  
 
==Vandermonde's Identity==
 
==Vandermonde's Identity==
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</asy>
 
</asy>
  
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This identity is known as the ''hockey-stick'' identity because, on Pascal's triangle, when the addends represented in the summation and the sum itself is highlighted, a hockey-stick shape is revealed. We can also flip the hockey stick because pascal's triangle is symettrical.
+
This identity is known as the ''hockey-stick'' identity because, on Pascal's triangle, when the addends represented in the summation and the sum itself is highlighted, a hockey-stick shape is revealed. We can also flip the hockey stick because pascal's triangle is symmetrical.
  
  
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'''Algebraic Proof 3'''
 
'''Algebraic Proof 3'''
  
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Consider the number of solutions to the equation <math>a_1</math>+<math>a_2</math>+<math>a_3</math>+<math>a_4</math>+<math>a_5</math>+<math>a_6</math>+.......<math>a_n-1</math>+<math>a_n</math> = N where each <math>a_i</math> is a  
+
Consider the number of solutions to the equation <math>a_1</math>+<math>a_2</math>+<math>a_3</math>+<math>a_4</math>+<math>a_5</math>+<math>a_6</math>+.......+<math>a_r</math> ≤ N where each <math>a_i</math> is a  
−
non-negative integer for 1≤i≤n.
+
non-negative integer for 1≤i≤r.
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=====METHOD 1=====
+
 
 +
'''METHOD 1'''
 
We know since all numbers on LHS are non-negative therefore 0≤N and N is a integer.
 
We know since all numbers on LHS are non-negative therefore 0≤N and N is a integer.
 
   
 
   
−
Therfore, <math>a_1</math>+<math>a_2</math>+<math>a_3</math>+<math>a_4</math>+<math>a_5</math>+<math>a_6</math>+.......+<math>a_n</math> = 0,1,2......N. Consider each case seperately.
+
Therfore, <math>a_1</math>+<math>a_2</math>+<math>a_3</math>+<math>a_4</math>+<math>a_5</math>+<math>a_6</math>+.......+<math>a_r</math> = 0,1,2......N. Consider each case seperately.
 +
 
 +
<math>a_1</math>+<math>a_2</math>+<math>a_3</math>+<math>a_4</math>+<math>a_5</math>+<math>a_6</math>+.......+<math>a_r</math> =0 by [[Stars-and-bars]] the equation has <math>\binom {0+r-1}{r-1}</math> solutions.
 +
 
 +
<math>a_1</math>+<math>a_2</math>+<math>a_3</math>+<math>a_4</math>+<math>a_5</math>+<math>a_6</math>+.......+<math>a_r</math> =1 by [[Stars-and-bars]] the equation has <math>\binom {1+r-1}{r-1}</math> solutions.
 +
 
 +
<math>a_1</math>+<math>a_2</math>+<math>a_3</math>+<math>a_4</math>+<math>a_5</math>+<math>a_6</math>+.......+<math>a_r</math> =2 by [[Stars-and-bars]] the equation has <math>\binom {2+r-1}{r-1}</math> solutions.
 +
 
 +
...........
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<math>a_1</math>+<math>a_2</math>+<math>a_3</math>+<math>a_4</math>+<math>a_5</math>+<math>a_6</math>+.......+<math>a_r</math> =N by [[Stars-and-bars]] the equation has <math>\binom {N+r-1}{r-1}</math> solutions.
 +
 
 +
Hence, the equation has <math>\binom {r-1}{r-1}</math>+ <math>\binom {r}{r-1}</math>+ <math>\binom {r+1}{r-1}</math>+.... <math>\binom {N+r-1}{r-1}</math>= <math>\sum^k_{i=r-1}{i\choose r-1}</math> (where k=N+r-1) SOLUTIONS.
 +
 
 +
'''METHOD 2'''
 +
Since, <math>a_1</math>+<math>a_2</math>+<math>a_3</math>+<math>a_4</math>+<math>a_5</math>+<math>a_6</math>+.......+<math>a_r</math> ≤ N Therefore we may say <math>a_1</math>+<math>a_2</math>+<math>a_3</math>+<math>a_4</math>+<math>a_5</math>+<math>a_6</math>+.......+<math>a_r</math> = N -m  where m is another non-negative integer.
 +
0 ≤<math>a_1</math>+<math>a_2</math>+<math>a_3</math>+<math>a_4</math>+<math>a_5</math>+<math>a_6</math>+.......+<math>a_r</math> ⇒ 0≤ N-m ⇒ m≤ N  So, we need not count this as an extra restriction.
 +
 
 +
Now, <math>a_1</math>+<math>a_2</math>+<math>a_3</math>+<math>a_4</math>+<math>a_5</math>+<math>a_6</math>+.......+<math>a_r</math> +m = N. Again by [[Stars-and-bars]] this has <math>\binom {N+r}{r}</math> solutions.
 +
 
 +
Therefore, the equation has <math>\binom {N+r}{r}</math> = <math>\binom {k+1}{r}</math>  solutions(As N+r-1 =k).  
  
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<math>a_1</math>+<math>a_2</math>+<math>a_3</math>+<math>a_4</math>+<math>a_5</math>+<math>a_6</math>+.......+<math>a_n</math> =0 by [[Stars-and-bars]] the equation has <cmath>\binom {0+n-1}n-1</cmath> solutions.
+
Since, both methods yeild the same answer ⇒''' <math>\sum^k_{i=r-1}{i\choose r-1}</math> = <math>\binom {k+1}{r}</math>'''. Taking r-1= p redirects to the honeystick identity.
 +
 
 +
~SANSGANKRSNGUPTA
  
 
==Another Identity==
 
==Another Identity==
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===Proof 2===
 
===Proof 2===
 
This is a special case of Vandermonde's identity, in which we set <math>m=n=r</math>.
 
This is a special case of Vandermonde's identity, in which we set <math>m=n=r</math>.
 +
 +
== Even Odd Identity ==
 +
<cmath>
 +
\sum_{k=0}^m (-1)^k \binom{n}{k}= (-1)^m \binom{n-1}{m}
 +
</cmath>
  
 
== Examples ==
 
== Examples ==
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* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_15 2021 AMC 12A  Problem 15]
 
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_15 2021 AMC 12A  Problem 15]
 
* [https://artofproblemsolving.com/wiki/index.php/1981_IMO_Problems 1981 IMO Problem 2]
 
* [https://artofproblemsolving.com/wiki/index.php/1981_IMO_Problems 1981 IMO Problem 2]
 +
* [https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_12  2022 AIME I Problem 12 ]
  
 
==See also==
 
==See also==

Latest revision as of 22:47, 27 October 2024

Pascal's Identity

Pascal's Identity states that

${n \choose k}={n-1\choose k-1}+{n-1\choose k}$

for any positive integers $k$ and $n$. Here, $\binom{n}{k}$ is the binomial coefficient $\binom{n}{k} = {}_nC_k = C_k^n$.

This result can be interpreted combinatorially as follows: the number of ways to choose $k$ things from $n$ things is equal to the number of ways to choose $k-1$ things from $n-1$ things added to the number of ways to choose $k$ things from $n-1$ things.

Proof

If $k > n$ then $\binom{n}{k} = 0 = \binom{n - 1}{k - 1} + \binom{n - 1}{k}$ and so the result is trivial. So assume $k \leq n$. Then

\begin{eqnarray*}\binom{n-1}{k-1}+\binom{n-1}{k}&=&\frac{(n-1)!}{(k-1)!(n-k)!}+\frac{(n-1)!}{k!(n-k-1)!}\\ &=&(n-1)!\left(\frac{k}{k!(n-k)!}+\frac{n-k}{k!(n-k)!}\right)\\ &=&(n-1)!\cdot \frac{n}{k!(n-k)!}\\ &=&\frac{n!}{k!(n-k)!}\\ &=&\binom{n}{k}. \qquad\qquad\square\end{eqnarray*}

Alternate Proofs

Here, we prove this using committee forming.

Consider picking one fixed object out of $n$ objects. Then, we can choose $k$ objects including that one in $\binom{n-1}{k-1}$ ways.

Because our final group of objects either contains the specified one or doesn't, we can choose the group in $\binom{n-1}{k-1}+\binom{n-1}{k}$ ways.

But we already know they can be picked in $\binom{n}{k}$ ways, so

\[{n \choose k}={n-1\choose k-1}+{n-1\choose k}\]


Also, we can look at Pascal's Triangle to see why this is. If we were to extend Pascal's Triangle to row n, we would see the term $\binom{n}{k}$. Above that, we would see the terms ${n-1\choose k-1}$ and ${n-1\choose k}$. Due to the definition of Pascal's Triangle, ${n \choose k}={n-1\choose k-1}+{n-1\choose k}$.

Vandermonde's Identity

Vandermonde's Identity states that $\sum_{k=0}^r\binom mk\binom n{r-k}=\binom{m+n}r$, which can be proven combinatorially by noting that any combination of $r$ objects from a group of $m+n$ objects must have some $0\le k\le r$ objects from group $m$ and the remaining from group $n$.

Video Proof

https://www.youtube.com/watch?v=u1fktz9U9ig

Combinatorial Proof

Think of the right hand side as picking $r$ people from $m$ men and $n$ women. Think of the left hand side as picking $k$ men from the $m$ total men and picking $r-k$ women from the $n$ total women. The left hand side and right hand side are the same, thus Vandermonde's identity must be true.

~avn

Algebraic proof

For all $x,$ \[(1+x)^m(1+x)^n = (1+x)^{m+n}\] The coefficients of $x^r$ on both sides must be the same, so using the Binomial Theorem, \[\sum_{k=0}^r \binom mk \binom n{r-k} = \binom {m+n}r.\]

Hockey-Stick Identity

For $n,r\in\mathbb{N}, n>r,\sum^n_{i=r}{i\choose r}={n+1\choose r+1}$.

[asy] int chew(int n,int r){ int res=1; for(int i=0;i<r;++i){ res=quotient(res*(n-i),i+1); } return res; } for(int n=0;n<9;++n){ for(int i=0;i<=n;++i){ if((i==2 && n<8)||(i==3 && n==8)){ if(n==8){label(string(chew(n,i)),(11+n/2-i,-n),p=red+2.5);} else{label(string(chew(n,i)),(11+n/2-i,-n),p=blue+2);} } else{ label(string(chew(n,i)),(11+n/2-i,-n)); } } } [/asy]

This identity is known as the hockey-stick identity because, on Pascal's triangle, when the addends represented in the summation and the sum itself is highlighted, a hockey-stick shape is revealed. We can also flip the hockey stick because pascal's triangle is symmetrical.


Proof

Inductive Proof

This identity can be proven by induction on $n$.

Base Case Let $n=r$.

$\sum^n_{i=r}{i\choose r}=\sum^r_{i=r}{i\choose r}={r\choose r}=1={r+1\choose r+1}$.

Inductive Step Suppose, for some $k\in\mathbb{N}, k>r$, $\sum^k_{i=r}{i\choose r}={k+1\choose r+1}$. Then $\sum^{k+1}_{i=r}{i\choose r}=\left(\sum^k_{i=r}{i\choose r}\right)+{k+1\choose r}={k+1\choose r+1}+{k+1\choose r}={k+2\choose r+1}$.

Algebraic Proof

It can also be proven algebraically with Pascal's Identity, ${n \choose k}={n-1\choose k-1}+{n-1\choose k}$. Note that

${r \choose r}+{r+1 \choose r}+{r+2 \choose r}+\cdots+{r+a \choose r}$ $={r+1 \choose r+1}+{r+1 \choose r}+{r+2 \choose r}+\cdots+{r+a \choose r}$ $={r+2 \choose r+1}+{r+2 \choose r}+\cdots+{r+a \choose r}=\cdots={r+a \choose r+1}+{r+a \choose r}={r+a+1 \choose r+1}$, which is equivalent to the desired result.

Combinatorial Proof 1

Imagine that we are distributing $n$ indistinguishable candies to $k$ distinguishable children. By a direct application of Balls and Urns, there are ${n+k-1\choose k-1}$ ways to do this. Alternatively, we can first give $0\le i\le n$ candies to the oldest child so that we are essentially giving $n-i$ candies to $k-1$ kids and again, with Balls and Urns, ${n+k-1\choose k-1}=\sum_{i=0}^n{n+k-2-i\choose k-2}$, which simplifies to the desired result.

Combinatorial Proof 2

We can form a committee of size $k+1$ from a group of $n+1$ people in ${{n+1}\choose{k+1}}$ ways. Now we hand out the numbers $1,2,3,\dots,n-k+1$ to $n-k+1$ of the $n+1$ people. We can divide this into $n-k+1$ disjoint cases. In general, in case $x$, $1\le x\le n-k+1$, person $x$ is on the committee and persons $1,2,3,\dots, x-1$ are not on the committee. This can be done in $\binom{n-x+1}{k}$ ways. Now we can sum the values of these $n-k+1$ disjoint cases, getting \[{{n+1}\choose {k+1}} ={{n}\choose{k}}+{{n-1}\choose{k}}+{{n-2}\choose{k}}+\hdots+{{k+1}\choose{k}}+{{k}\choose{k}}.\]


Algebraic Proof 2

Apply the finite geometric series formula to $(1+x)$: \[1+(1+x)+(1+x)^2+...+(1+x)^n=\frac{(1+x)^{n+1}-1}{(1+x)-1}\] Then expand with the Binomial Theorem and simplify: \[1+(1+x)+(1+2x+x^2)+...+ \left (\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+...+\binom{n}{n-1}x^{n-1}+\binom{n}{n}x^n \right )=\binom{n+1}{1}+\binom{n+1}{2}x+...+\binom{n+1}{n+1}x^n\] Finally, equate coefficients of $x^m$ on both sides: \[\binom{0}{m}+\binom{1}{m}+\binom{2}{m}+...+\binom{n}{m}=\binom{n+1}{m+1}\] Since for $i<m$, $\binom{i}{m}=0$, this simplifies to the hockey stick identity.

Algebraic Proof 3

Consider the number of solutions to the equation $a_1$+$a_2$+$a_3$+$a_4$+$a_5$+$a_6$+.......+$a_r$ ≤ N where each $a_i$ is a non-negative integer for 1≤i≤r.

METHOD 1 We know since all numbers on LHS are non-negative therefore 0≤N and N is a integer.

Therfore, $a_1$+$a_2$+$a_3$+$a_4$+$a_5$+$a_6$+.......+$a_r$ = 0,1,2......N. Consider each case seperately.

$a_1$+$a_2$+$a_3$+$a_4$+$a_5$+$a_6$+.......+$a_r$ =0 by Stars-and-bars the equation has $\binom {0+r-1}{r-1}$ solutions.

$a_1$+$a_2$+$a_3$+$a_4$+$a_5$+$a_6$+.......+$a_r$ =1 by Stars-and-bars the equation has $\binom {1+r-1}{r-1}$ solutions.

$a_1$+$a_2$+$a_3$+$a_4$+$a_5$+$a_6$+.......+$a_r$ =2 by Stars-and-bars the equation has $\binom {2+r-1}{r-1}$ solutions.

........... $a_1$+$a_2$+$a_3$+$a_4$+$a_5$+$a_6$+.......+$a_r$ =N by Stars-and-bars the equation has $\binom {N+r-1}{r-1}$ solutions.

Hence, the equation has $\binom {r-1}{r-1}$+ $\binom {r}{r-1}$+ $\binom {r+1}{r-1}$+.... $\binom {N+r-1}{r-1}$= $\sum^k_{i=r-1}{i\choose r-1}$ (where k=N+r-1) SOLUTIONS.

METHOD 2 Since, $a_1$+$a_2$+$a_3$+$a_4$+$a_5$+$a_6$+.......+$a_r$ ≤ N Therefore we may say $a_1$+$a_2$+$a_3$+$a_4$+$a_5$+$a_6$+.......+$a_r$ = N -m where m is another non-negative integer. 0 ≤$a_1$+$a_2$+$a_3$+$a_4$+$a_5$+$a_6$+.......+$a_r$ ⇒ 0≤ N-m ⇒ m≤ N So, we need not count this as an extra restriction.

Now, $a_1$+$a_2$+$a_3$+$a_4$+$a_5$+$a_6$+.......+$a_r$ +m = N. Again by Stars-and-bars this has $\binom {N+r}{r}$ solutions.

Therefore, the equation has $\binom {N+r}{r}$ = $\binom {k+1}{r}$ solutions(As N+r-1 =k).

Since, both methods yeild the same answer ⇒ $\sum^k_{i=r-1}{i\choose r-1}$ = $\binom {k+1}{r}$. Taking r-1= p redirects to the honeystick identity.

~SANSGANKRSNGUPTA

Another Identity

\[\sum_{i=0}^k \binom{k}{i}^2=\binom{2k}{k}\]

Hat Proof

We have $2k$ different hats. We split them into two groups, each with k hats: then we choose $i$ hats from the first group and $k-i$ hats from the second group. This may be done in $\binom{k}{i}^2$ ways. Evidently, to generate all possible choices of $k$ hats from the $2k$ hats, we must choose $i=0,1,\cdots,k$ hats from the first $k$ and the remaining $k-i$ hats from the second $k$; the sum over all such $i$ is the number of ways of choosing $k$ hats from $2k$. Therefore $\sum_{i=0}^k \binom{k}{i}^2=\binom{2k}{k}$, as desired.

Proof 2

This is a special case of Vandermonde's identity, in which we set $m=n=r$.

Even Odd Identity

\[\sum_{k=0}^m (-1)^k \binom{n}{k}= (-1)^m \binom{n-1}{m}\]

Examples

See also

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