Shi Jin MATH 3313 Fall 2011 2.6 Principle of Counting 21. a) Prove Let n be a positive integer and r be an integer such that 0≤r≤n. n # &
n r n *r
n
(a) For a, b " R,(a + b) = )% (a b . r
r =0 $ '
n " %
n
n
(a
+
b)
=
i) P(n): ($# r '&arbn )r -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐open statement depends on n !
r =0
P(1) is true-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐basic step !
"1 %
"1%
(a + b)1 = $ 'a 0b1(0 + $ 'a1b1(1 #0 &
#1&
a+b=1*1*b+1*a*1 !
a+b=b+a-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐True ii) Assume P(n) is true and show P(n+1) is also true-‐-‐-‐-‐-‐-‐induction step n " %
n r n )r
n
Assume: (a + b) = ($ 'a b r
r =0 # &
n +1
W.T.S: (a + b)
!
n +1
" n +1% r n +1)r
= ($
'a b
r &
r =0 #
Start from the right hand side !
n +1
" n +1% r n +1)r
'a b
r &
r =0
($#
n
=b
n +1
=b
n +1
!
" n +1% r n +1)r
+ ($
+ a n +1 'a b
r
#
&
r =1
"" n % " n %%
r n +1(r
+ )$$ ' + $
+ a n +1
''a b
r
r (1&&
r =1 ## & #
-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐Property of Binomial Coefficients
n
!
!
n
=b
n +1
" n % r n +1)r n " n % r n +1)r
+ ($ 'a b
+ ($
+ a n +1 'a b
r
r )1&
r =1 # &
r =1 #
!
n " %
"
%
n r n +1)r % " n " n % r n +1)r
n
+ a n * a' ' + $ ($
= $b * b + ($ 'a b
'a b
r
#
& # r =1 # r )1&
&
r =1 # &
!
n " %
"
%
n r n )r % " n " n % r )1 n +r )1
n
+ a n ' = b$ b + ($ 'a b ' + a$($
'a b
r
#
& # r =1 # r )1&
&
r =1 # &
!
n " %
"
%
n r n )r % " n )1 " n % r n )r
n
n
= b$ b + ($ 'a b ' + a$($ 'a b + a ' r
#
& # r =0 # r &
&
r =1 # &
!
n " %
"" n %
n r n )r % " n )1 " n % r n )r " n % n 0 %
0 n
= b$$ 'a b + ($ 'a b ' + a$ ($ 'a b + $ 'a b ' r
# n&
## 0 &
& # r =0 # r &
&
r =1 # &
!
" n " n%
% " n " n%
%
r n )r
= b$ ($ 'a b ' + a$($ 'a rb n )r ' # r =0 # r &
& # r =0 # r &
&
!
" n " n%
%
= ( a + b) $ ($ 'a rb n )r ' # r =0 # r &
&
1
1
= ( a + b) (a + b) n !
= (a + b)
n +1
= (a + b) n +1 !
b) Prove !
" n % " n (1% " n (1%
(c) $ ' = $
' +$
' for r≥1. # r & # r & # r (1&
!
" n%
n!
We know that $ ' =
# r & r!(n ( r)!
Start from right hand side # n "1!
& # n "1&
%
( +%
( $ r ' $ r "1'
!
=
(n "1)!
(n "1)!
+
r!(n "1 " r)! (r "1)!((n "1) " (r "1))!
=
(n "1)!
(n "1)!
+
r!(n "1 " r)! (r "1)!(n " r)!
=
(n "1)!
(n "1)!
+
r(r "1)!(n "1 " r)! (r "1)!(n " r)(n "1 " r)!
=
#1
(n "1)!
1 &
*% +
( (r "1)!(n "1 " r)! $ r (n " r) '
=
# (n " r) + r &
(n "1)!
*%
( (r "1)!(n "1 " r)! $ r(n " r) '
=
# n &
(n "1)!
*%
( (r "1)!(n "1 " r)! $ r(n " r) '
=
n * (n "1)!
r * (r "1)!*(n " r) * (n "1 " r)!
=
n!
n!
=
r!(n " r)! r!(n " r)!
!
!
!
!
!
!
!
23.a) Give a combinatorial proof that if n is an odd integer, then the number of ways to !
select an even number of objects from a set of n objects is equal to the number of ways to select an odd number of objects. Example: N=5 and odd number Number of ways to select odd number of objects = Number of ways to select even "5 % "5 % "5 % "5 % "5 % " 5 %
$ ' + $ ' + $ ' = $ ' + $ ' + $ ' #5 & # 3& #1 & #0 & #2 & # 4 &
1+10+5=1+10+5 !
Proof: Suppose I select an odd number of objects (r) from n " n%
$ ' ways # r&
!
then I select the remaining objects from n # n &
%
( ways $ n " r'
because of the complement !
by complement I mean A={a,b,c,d,e} "5%
$ ' = {(a),(b),(c),(d),(e)} #1&
!
" 5%
$ ' = {(b,c,d,e),(a,c,d,e),(a,b,d,e),(a,b,c,e),(a,b,c,d)} # 4&
!
" n% " n %
therefore $ ' = $
' # r & # n ( r&
So count every even number and every odd number from 0 to n. we would find that k !
possibilities of even number as well as k possibilities of odd number since each one has a complement we are guaranteed that the number of ways to select an even number of objects from a set of n objects is equal to the number of ways to select an odd number of objects. Note: if N=even=4 " 4% " 4% " 4% " 4% " 4%
" 4%
$ ' + $ ' + $ ' + $ ' + $ ' -‐-‐-‐-‐-‐each one has a pair of complement except $ ' its complement # 0 & # 1 & # 2 & # 3& # 4 &
# 2&
is itself. Since we don’t write it twice therefore it remain alone. !
b) Give a combinatorial proof of Vandermonde’s identity: For !positive integers m and n, and r an integer such that 0≤r≤n+m. " n + m% " n %" m% " n %" m % " n %" m %
" n %" m%
$
' = $ '$ ' + $ '$
' + $ '$
' + ...+ $ '$ ' # r & # 0 &# r & # 1 &# r (1& # 2 &# r ( 2&
# r &# 0 &
!
Let A be a set with n elements |A|=n B be a set with m elements |B|=m Consider A and B are disjoint (no elements in common) then by sum’s rule theorem 2.6.1 |AUB|=n+m " n + m%
$
' = Number of subsets with r elements from the set with size (n+m) # r &
Example: so start with n=2, m=3, and r=4 using the above formula we get !
0 elements from A and 4 elements from B 1 element from A and 3 elements from B 2 elements from A and 2 elements from B 3 elements from A and 1 element from B 4 elements from A and 0 elements from B "2%" 3% " 2%" 3% "2%" 3% " 2%" 3% " 2 %" 3%
= $ '$ ' + $ '$ ' + $ '$ ' + $ '$ ' + $ '$ ' since they are disjoint we can add. #0&# 4 & # 1&# 3& #2&# 2& # 3& # 4 &# 0&
!
" n%
Proof: Assume a subset with k elements will be selected from A then there are $ ' ways #k&
# m &
Then we must select the remaining r-‐k elements from B then there are %
( ways ! $r " k'
" n% # m &
Therefore there are $ ' %
( ways to selected a set of size k # k & $r " k'
!
r " %"
n m %
If we consider all values of k we get a summation of )$ '$
' k
r
(
k
#
&
#
&
k
=0
! !
"n %" m % " n %" m % "n %" m %
" n %" m %
= $ '$ ' + $ '$
' + $ '$
' + ...+ $ '$ '
#0 &# r & # 1 &#r (1& #2 &# r ( 2 &
# r &# 0 & !
!
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