Pascal's Identity states that
for any positive integers and . Here, is the binomial coefficient .
This result can be interpreted combinatorially as follows: the number of ways to choose things from things is equal to the number of ways to choose things from things added to the number of ways to choose things from things.
If then and so the result is trivial. So assume . Then
Here, we prove this using committee forming.
Consider picking one fixed object out of objects. Then, we can choose objects including that one in ways.
Because our final group of objects either contains the specified one or doesn't, we can choose the group in ways.
But we already know they can be picked in ways, so
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 . Above that, we would see the terms and . Due to the definition of Pascal's Triangle, .
Vandermonde's Identity states that , which can be proven combinatorially by noting that any combination of objects from a group of objects must have some objects from group and the remaining from group .
Think of the right hand side as picking people from men and women. Think of the left hand side as picking men from the total men and picking women from the total women. The left hand side and right hand side are the same, thus Vandermonde's identity must be true.
For all The coefficients of on both sides must be the same, so using the Binomial Theorem,
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 can be proven by induction on .
Base Case Let .
Inductive Step Suppose, for some , . Then .
It can also be proven algebraically with Pascal's Identity, . Note that
, which is equivalent to the desired result.
Combinatorial Proof 1
Imagine that we are distributing indistinguishable candies to distinguishable children. By a direct application of Balls and Urns, there are ways to do this. Alternatively, we can first give candies to the oldest child so that we are essentially giving candies to kids and again, with Balls and Urns, , which simplifies to the desired result.
Combinatorial Proof 2
We can form a committee of size from a group of people in ways. Now we hand out the numbers to of the people. We can divide this into disjoint cases. In general, in case , , person is on the committee and persons are not on the committee. This can be done in ways. Now we can sum the values of these disjoint cases, getting
Algebraic Proof 2
Apply the finite geometric series formula to : Then expand with the Binomial Theorem and simplify: Finally, equate coefficients of on both sides: Since for , , this simplifies to the hockey stick identity.
Algebraic Proof 3
Consider the number of solutions to the equation ++++++.......+ ≤ N where each 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, ++++++.......+ = 0,1,2......N. Consider each case seperately.
++++++.......+ =0 by Stars-and-bars the equation has solutions.
++++++.......+ =1 by Stars-and-bars the equation has solutions.
++++++.......+ =2 by Stars-and-bars the equation has solutions.
........... ++++++.......+ =N by Stars-and-bars the equation has solutions.
Hence, the equation has + + +.... = (where k=N+r-1) SOLUTIONS.
METHOD 2 Since, ++++++.......+ ≤ N Therefore we may say ++++++.......+ = N -m where m is another non-negative integer. 0 ≤++++++.......+ ⇒ 0≤ N-m ⇒ m≤ N So, we need not count this as an extra restriction.
Now, ++++++.......+ +m = N. Again by Stars-and-bars this has solutions.
Therefore, the equation has = solutions(As N+r-1 =k).
Since, both methods yeild the same answer ⇒ = . Taking r-1= p redirects to the honeystick identity.
We have different hats. We split them into two groups, each with k hats: then we choose hats from the first group and hats from the second group. This may be done in ways. Evidently, to generate all possible choices of hats from the hats, we must choose hats from the first and the remaining hats from the second ; the sum over all such is the number of ways of choosing hats from . Therefore , as desired.
This is a special case of Vandermonde's identity, in which we set .
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