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Discrete Mathematics
Ch. 6 Counting and Probability
Today we will review sections 6.6, 6.7
Instructor: Hayk Melikyan
[email protected]
Melikyan/DM/Fall09
1
Algebra of Combinations and Pascal’s Triangle
The number of r-combinations from a set of n elements equals
the number of (n – r) combinations from the same set.
C(n, r) = C(n, n-r)
Melikyan/DM/Fall09
2
Pascal's Triangle:
Theorem: (Pascal’s Formula). Let n and r be
positive integers and r  n, then
C(n + 1, r) = C(n, r – 1) + C(n, r)
Proof: Algebraic verses Combinatorial
A combinatorial proof is an argument that establishes an algebraic
fact by relying on counting principles.
Melikyan/DM/Fall09
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Proof (Combinatorial Version)
Is S is an set with n+1 elements. The number of all subsets
of size r can be calculated as follows:
Melikyan/DM/Fall09
4
Pascal’s Triangle
Melikyan/DM/Fall09
5
Exercises
Show that:
1 * 2 + 2 * 3 + n * (n + 1) = 2 * C(n + 2, 3)
Prove that :
C(n, 0)2 + C(n, 1)2 + … + C(n, n)2 = C(2n, n)
Melikyan/DM/Fall09
6
Binomial Formula
Theorem (Binomial Theorem): Given any real numbers a
and b and any nonnegative integer n
(a + b)n =
= an + C(n, 1)an-1b + C(n, 2)an-2b2 + . . . + C(n, n-1)an-1b + bn
=
=
n
 C ( n, k ) a
n k
bk
k 0
Melikyan/DM/Fall09
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Example:
n
1. Show that  C(n, k ) = 2n
k 0
n
 ( 1)
2. Show that
k
C ( n, k ) = 0
k 0
n
k
3. Express  kC(n, k )3 in the closed form
k 1
Melikyan/DM/Fall09
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What about ?
Melikyan/DM/Fall09
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