Download 3.3 Combinations of sets

 4.5.2 Applications of Inclusion-Exclusion
principle
 The number of r-combinations of multiset S
 1. The number of r-combinations of multiset S
 If r<n, and there is, in general, no simple
formula for the number of r-combinations of
S. Nonetheless a solution can be obtained by
the inclusion-exclusion principle 4.5 .
 Example: Determine the number of 10-combinations of
multiset S={3·a,4·b,5·c}.
 Solution：We shall apply the inclusion-exclusion principle
to the set Y of all 10-combinations of the multiset D={·a,
·b, ·c}.
 Let P1 be the property that a 10-combination of D has more
than 3 a’s. Let P2 be the property that a 10-combination of
D has mote than 4 b’s. Let P3 be the property that a 10combination of D has mote than 5 c’s.
 For i=1,2,3 let Ai be the set consisting of those 10combinations of D which have property Pi.
 The number of 10-combinations of S is then the number of
10-combinations of D which have none of the properties P1,
P2, and P3.
A1  A2  A3
 The set A1 consists of all 10-combinations of D
in which a occurs at least 4 time.
 If we take any one of these 10-combinations in
A1 and remove 4 a’s, we are left with a 6combination of D.
 Conversely, if we take a 6-combination of D
and add 4 a’s to it, we get a 10-combination of
D in which a occurs at least 4 times.
 Thus the number of 10-combinations in A1
equals the number of 6-combinations of D.
Hence, |A1|=C(3+6-1,6)=C(8,6)=C(8,2)，
 Example: What is the number of integeal solutions
of the equation
 x1+x2+x3=5
 which satisfy 0x12,0x22,1x35?
 Solution: We introduce new variables,
 x3'=x3-1
 and our equation becomes x1+x2+x3'=4.
 The inequalities on the xi and x3' are satisfied if and
only if
 0x12,0x22, 0x3'4.
 4-combinations of multiset {2·a,2·b,4·c}
 2.Derangements
 A derangement of {1,2,…,n} is a permutation
i1i2…in of {1,2,…,n} in which no integer is in
its natural position:
 i11,i22,…,inn.
 We denote by Dn the number of derangements
of {1,2,…,n}.
 Theorem 4.15：For n1,
Dn  n! (1 
1 1 1
1
     ( 1) n )
1! 2! 3!
n!
 Proof: Let S={1,2,…,n} and X be the set of all
permutations of S. Then |X|=n!.
 For j=1,2,…,n, let pj be the property that in a
permutation, j is in its natural position. Thus
the permutation i1,i2,…,in of S has property pj
provided ij=j. A permutation of S is a
derangement if and only if it has none of the
properties p1,p2,…,pn.
 Let Aj denote the set of permutations of S
with property pj ( j=1,2,…,n).
 Example:(1)Determine the number of
permutations of {1,2,3,4,5,6,7,8,9} in which no
odd integer is in its natural position and all
even integers are in their natural position.
 (2) Determine the number of permutations of
{1,2,3,4,5,6,7,8,9} in which four integers are in
their natural position.
 3. Permutations with relative forbidden
position
 A Permutations of {1,2,…,n} with relative
forbidden position is a permutation in which
none of the patterns i,i+1(i=1,2,…,n) occurs.
We denote by Qn the number of the
permutations of {1,2,…,n} with relative
forbidden position.
 Theorem 4.16：For n1,
 Qn=n!-C(n-1,1)(n-1)!+C(n-1,2)(n-2)!-…+(-1)n-1
C(n-1,n-1)1!
 Proof: Let S={1,2,…,n} and X be the set of all
permutations of S. Then |X|=n!.
 j(j+1), pj(1,2,…,n-1)
 Aj: pj
 Qn=Dn+Dn-1
4.6 Generating functions
 4.6.1 Generating functions
 Let S={n1•a1,n2•a2,…,nk•ak}, and n=n1+n2+…+nk=|S|，then
the number N of r-combinations of S equals
 (1)0 when r>n
 (2)1 when r=n
 (3) N=C(k+r-1,r) when ni r for each i=1,2,…,n.
 (4)If r<n, and there is, in general, no simple formula for the
number of r-combinations of S.
 A solution can be obtained by the inclusion-exclusion
principle and technique of generating functions.
 6-combination a1a1a3a3a3a4
 xi1xi2…xik= xi1+i2+…+ik=xr
 r-combination of S
 Definition 1: The generating function for the
sequence a0,a1,…,an,… of real numbers is the
infinite series f(x)=a0+a1x+a2x2+…+anxn+…,
and


i
a
x

b
x
 i i
i
i 0
i 0
if only if ai=bi for all i=0,1, …n, …
 We can define generating function for finite
sequences of real numbers by extending a
finite sequences a0,a1,…,an into an infinite
sequence by setting an+1=0, an+2=0, and so on.
 The generating function f(x) of this infinite
sequence {an} is a polynomial of degree n
since no terms of the form ajxj, with j>n occur,
that is f(x)=a0+a1x+a2x2+…+anxn.
 Example: (1)Determine the number of ways in which postage
of r cents can be pasted on an envelope using 1 1-cent,1 2-cent,
1 4-cent, 1 8-cent and 1 16-cent stamps.
 (2)Determine the number of ways in which postage of r cents
can be pasted on an envelope using 2 1-cent, 3 2-cent and 2 5cent stamps.
 Assume that the order the stamps are pasted on does not
matter.
 Let ar be the number of ways in which postage of r cents.
Then the generating function f(x) of this sequence {ar} is
 (1)f(x)=(1+x)(1+x2)(1+x4)(1+x8)(1+x16)
 (2)f(x)=(1+x+x2)(1+x2+(x2)2+(x2)3)(1+x5+(x5)2)）
 =1+x+2x2+x3+2x4+2x5+3x6+3x7+2x8+2x9+2x10+3x11
+3x12+2x13+ 2x14+x15+2x16+x17+x18。
 Example: Use generating functions to
determine the number of r-combinations of
multiset S={·a1,·a2,…, ·ak }.
 Solution: Let br be the number of rcombinations of multiset S. And let generating
functions of {br} be f(y)，
 (1+y+y2+…)k=? f(y)

1
r
f ( y) 

C
(
k

r

1
,
r
)
y

(1  y ) k r  0
 Example: Use generating functions to determine the
r
r
number
of
multiset
C (kof
 r-combinations

1
,
)
r

2
n
(
n
 0,1,)
2
2
ar   ·a ,n ·a ,…,n ·a }.
S={n
rk  2k n  1(n  0,1,)
1 0 1 2 2
 Solution: Let generating functions of {br} be f(y)，
 f(y)=(1+y+y2+…+yn1)(1+y+y2+…+yn2)…(1+y+y2+…+ynk)
 Example: Let S={·a1,·a2,…,·ak}. Determine the
number of r-combinations of S so that each of the k
types of objects occurs even times.
 Solution: Let generating functions of {br} be f(y)，
 f(y)=(1+y2+y4+…)k=1/(1-y2)k
 =1+ky2+C(k+1,2)y4+…+C(k+n-1,n)y2n+…
 Example: Determine the number of 10combinations of multiset S={3·a,4·b,5·c}.
 Solution: Let generating functions of {ar} be
f(y)，
 f(y)=(1+y+y2+y3)(1+y+y2+y3+y4)(1+y+y2+y3+y4
+y5)
 =1+3y+6y2+10y3+14y4+17y5+18y6+17y7+14y8+
10y9+6y10+3y11+y12
 Example: What is the number of integral
solutions of the equation
 x1+x2+x3=5
 which satisfy 0x1,0x2,1x3?
 Let x3'=x3-1，
 x1+x2+x3'=4, where 0x1,0x2,0x3'
f ( y )  (1  y  y 2  )(1  y  y 2  )(1  y  y 2  )

1
r


C
(
r

2
,
r
)
y

(1  y )3 r  0
 Next: Exponential Generating functions;
 Recurrence Relations P13, P112 (Sixth) OR P13,P100(Fifth)
 Exercise: 1. Determine the number of 12-combinations of the multiset
S={4·a,3·b,5·c, 4·d }.
 2. Determine the number of solutions of the equation x1+x2+x3+x4=14 in
nonnegative integers x1,x2,x3, and x4 not exceeding 8.
 3.Determine the number of permutations of {1,2,3,4,5,6,7,8} in which no
even integer is in its natural position.
 4.Determine the number of permutations of {1,2,…,n} in which exactly k
integers are in their natural positions.
 5.Eight boys are seated around a carousel. In how many ways can they
change seats so that each has a different boy in front of him?
 6.Let S be the multiset {·e1,·e2,…, ·ek}. Determine the generating
function for the sequence a0, a1, …,an, … where an is the number of ncombinations of S with the added restriction:
 1) Each ei occurs an odd number of times.
 2) the element e2 does not occur, and e1 occurs at most once.
 7.Determine the generating function for the number an of nonnegative
integral solutions of 2e1+5e2+e3+7e4=n