Discrete Mathematics University of Kentucky CS 275 Spring ... - MGNet

Theorem: The number of r-combinations of a set with n elements with n,r,N 0 is
C(n,r) = n # r
!
"
$
.
%
&
Proof: The r-permutations can be formed using C(n,r) r-combinations and then
ordering each r-combination, which can be done in P(r,r) ways. So,
P(n,r) = C(n,r)$P(r,r)
or
C(n,r) = P(r,r)
P(n,r) = n!
(n-r)! !(r-r)! = n!
r! r!(n-r)! .
Theorem: C(n,r) = C(n,n-r) for 0;r;n.
Definition: A combinatorial proof of an identity is a proof that uses counting
arguments to prove that both sides f the identity count the same objects, but in
different ways.
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Binomial Theorem: Let x and y be variables. Then for n,N,
(x+y) n n ! n$
= # &
' x
j=0#
j
n-j y j .
&
Proof: Expanding the terms in the product all are of the form x n-j y j for
j=0,1,…,n. To count the number of terms for x n-j y j , note that we have to choose
n-j x’s from the n sums so that the other j terms in the product are y’s. Hence,
the coefficient for x n-j y j ! n $
is # &
# n-j
= ! n $
# &
& # j
.
&
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Example: What is the coefficient of x 12 y 13 in (x+y) 25 !
? 25 $
# &
#
13
= 5,200,300.
&
n ! n$
Corollary: Let n,N 0 . Then # &
' k=0#
k
= 2n .
&
Proof: 2 n = (1+1) n n ! n$
= # &
# k
1k 1 n-k n ! n$
= # &
'
&
' .
k=0
k=0#
k&
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