PEU5302 Combinatorics - Combinatorial Proofs
Dr. M.F.M. Sabri
Department of Mathematics
The Open University of Sri Lanka
Dr. M.F.M. Sabri (Department of Mathematics
PEU5302
The Open
Combinatorics
University of
- Combinatorial
Sri Lanka)
Proofs
1/6
Combinatorial Proofs
Combinatorial Proofs
Example
Prove that
n
Cr = n Cn−r .
Idea:
Here we use the proof technique called Double Counting. We argue what
count is given by the L.H.S. and we argue the same count in a different
way to obtain the R.H.S.
Dr. M.F.M. Sabri (Department of Mathematics
PEU5302
The Open
Combinatorics
University of
- Combinatorial
Sri Lanka)
Proofs
2/6
Combinatorial Proofs
Combinatorial proofs cont...
Proof.
Notice that n Cr is the number of ways to choose r number of objects from
n number of objects. Each such way of choosing r number of objects is
equivalent to choosing which n − r number of objects are remaining. So
the number of ways to choose r number of objects from n is equal to the
number of ways to choose n − r number of objects from n. So we have
n
Cr = n Cn−r .
Dr. M.F.M. Sabri (Department of Mathematics
PEU5302
The Open
Combinatorics
University of
- Combinatorial
Sri Lanka)
Proofs
3/6
Combinatorial Proofs
Combinatorial proofs cont...
Example
Show that
n
Cr = n−1 Cr −1 + n−1 Cr .
Proof.
Observe that n Cr is the number of ways to choose r objects from n
objects. Consider a special objects P. Then there are two cases. Either we
choose P or we don’t. In the first case P will be chosen in advance and
the remaining r − 1 objects will be selected from the remaining n − 1
objects. The number of ways to do that is n−1 Cr −1 . In the second case P
is not chosen. So we have to chose r number of objects anyways from the
remaining n − 1 number of objects. The number of ways to do that is
n−1 C . So the total number of ways to choose r objects from n objects is
r
n−1 C
n−1 C . Hence
r −1 +
r
n
Cr = n−1 Cr −1 + n−1 Cr .
Dr. M.F.M. Sabri (Department of Mathematics
PEU5302
The Open
Combinatorics
University of
- Combinatorial
Sri Lanka)
Proofs
4/6
Combinatorial Proofs
Combinatorial proofs cont...
Example
Can you argue why the following identity is true?
n
Cr = n−2 Cr −2 + 2 × n−2 Cr −1 + n−2 Cr
Dr. M.F.M. Sabri (Department of Mathematics
PEU5302
The Open
Combinatorics
University of
- Combinatorial
Sri Lanka)
Proofs
5/6
Combinatorial Proofs
Combinatorial proofs cont...
Example
Can you argue why the following identity is true?
n
Cr = n−2 Cr −2 + 2 × n−2 Cr −1 + n−2 Cr
Example
Can you argue why the following identity is true?
n
Cr r Ck = n Ck n−k Cr −k
Dr. M.F.M. Sabri (Department of Mathematics
PEU5302
The Open
Combinatorics
University of
- Combinatorial
Sri Lanka)
Proofs
5/6
Combinatorial Proofs
Combinatorial proofs cont...
Vandermonde’s Identity
m+n
Cr =
r
X
m
Ck × n Cr −k
k=0
Dr. M.F.M. Sabri (Department of Mathematics
PEU5302
The Open
Combinatorics
University of
- Combinatorial
Sri Lanka)
Proofs
6/6
Combinatorial Proofs
Combinatorial proofs cont...
Vandermonde’s Identity
m+n
Cr =
r
X
m
Ck × n Cr −k
k=0
Hockey-Stick Identity
n+1
Cr +1 = n Cr + n−1 Cr + n−2 Cr + · · · + r +1 Cr + r Cr
Dr. M.F.M. Sabri (Department of Mathematics
PEU5302
The Open
Combinatorics
University of
- Combinatorial
Sri Lanka)
Proofs
6/6