Inclusion and exclusion principle; inversion

formulaeThm 10.1. Let S be an N-set; Ei’s, i=1..r, be subsets of S. For any subset M of [r], define N(M) = | iM Ei| and for j=0..r, define Nj= |M|=j N(M). Then the number of elements of S not in any Ei is N-N1+N2-N3+…+(-1)r Nr. (*)

Pf: If x is in S and not in any Ei, then x contributes 1 to (*).If x is in S and in exactly k of Ei, then the contribution to (*) is1 ... ( 1) (1 1) 0.

1 2k kk k k

k

p2.p2.

Eg. 10.1. Let dn denote the number of permutations of 1, 2,…, n such that (i) i. Let Ei be the subset of those permutations with (i)=i. By Thm 10.1 we have:

10 0

00

0 0

11 2

( 1)( 1) ( )! ! !/ . ( 1) .

!

Consider ( ) : , 1. If ( ) : ( ), then!

( ) ( )( )! !

1 ( ...

! ( 1)! ( 2)!2! 1

in ni n

n n ni i

nx

nn

m n

nm n

m

nd n i n n e d nd

i i

xD x d d F x e D x

n

x xF x d

m n

dd d

m m m

1

1

0 0

0

0

(

)!( 1)! ( )!

)

( ) (1 )

(1 ) .!

.

m

m rr

x

mm

m

mm

m m

md

r

D x e x

dx

m m

xx x

m

p3.p3.

Eg. Consider a set Z with n blue points and m red points. How many k-subsets of Z consist of red points only?

i

i

0

The answer is trivially

If we take S to be all the k-subsets of Z and E those

k-subsets that contains the i-th blue point, then

by Thm 10.1 we have:

(-1

if .

) ,n

i

mm k

k

n m n i

i k i

which is .

m

k

p4.p4.

i

0

1

0

0

The identity (-1) = can be

proved directly.

1Consider (1 ) ( )

( 1)( 2)...( ) = ( 1)

!

n

i

a j

j

j j

j

n m n i m

i k i k

ax x

j

a a a jx

j

0 0

( )! = .

! !j j

j j

a ja jx x

ja j

p5.p5.

0

1

i

is the coefficient of in the expansion

of (1 ) . So the lef

(

t-hand side of the above identity

is the coefficient of in the

-1) = .

e

k i

k

i

m n

k

n n m n i m

i k i k

m n i

kx

i

x

x

1xpansion of (1 ) .

If m k-1, it is 0 and if m k, it is .

k mx

m

k

Eg. (The Euler function)

1 21 2

i

Let .... be a positive integer.

Denote ( ) the number of integers with 1

such that the gcd( , ) 1. By Thm 10.1 and let E be

the set of integers in [ ] divisible by , 1 .

ra a ar

i

n p p p

n k k n

n k

n p i r

11 1

1( ) (1 ).

Thenr r

ii i j ri i j i

n nn n n

p p p p

p7.p7.

p8.p8.

d | n d’ | d f(d, d’) = d’ | n d | n/d’ f(dd’, d’)

Example: Take n=6.

0 1 2 3 4 5 6

6

5

43

2

1

d

d’

Example: Count the number of Nn of circular sequences of 0’s and 1’s, where two sequences obtained by a rotation are considered the same.

Sol: Let M(d) be the number of circular sequences of length d

that are not periodic. Then Nn = d|n M(d). Why? Observe that 2n = d|n dM(d). By Thm 10.4, nM(n) = d|n μ(d) 2n/d. Why? So, M(d) = (1/d) d’|d μ(d’) 2d/d’. Thus, Nn = d|n M(d) = d|n M(d) = d|n(1/d) d’|d μ(d’) 2d/d’

= d|n(1/d) d’|d μ(d/d’) 2d’

= d’|n d|n/d’ (1/dd’) μ(d) 2d’

= d’|n 2d’/d’ d| n/d’ μ(d)/d = d’|n 2d’/d’ (n/d’)/(n/d’) = (1/n) d’|n 2d’ (n/d’)

00

10

1

Example: How many ways can we seat N couples around a circular table so that no husband will sit on either side of his wife?

Sol: Let Ar be the number of ways that there are r couples

sitting together. Treat the r couples as a unit, then it turns out to

arrange 2N - r units at a circle. So Ar = 2r (2N-r-1)!.

By Thm 10.1, there are r=0..N (-1)r C(N, r) Ar ways to seat N couples around a circular table so that no couple sit together.

Example: (Menage Probleme, by Lucas 1891) How many ways can we seat N couples at a circular table so that men and women are in alternate places and no couple sit next to each

other?Sol: Call the ladies 1 to N and the corresponding men also

1 to N. Assume the women have been seated at alternate places.

Let Ar be the ways to seat r husband incorrectly. Consider a circular sequence of 2N positions. Put a 1

in position 2i-1 if husband i is sitting to the right of his wife; put a 1 in position 2i if husband i is sitting to the left of his wife. Put zeroes in the remaining positions.

Ar is the number of 0-1 circular sequences with exactly r 1’s without two adjacent. Is 0110, possible?

Let A’r be the number of sequences

starting with a 1 (followed by a 0). By considering 10 as one symbol, we must choose r-1 positions out of 2N-r-1 positions.

1

0

2i-1

2i

01

To count the number A’’r of sequences starting with a 0, place the 0 at the end, and then it amounts to choosing r out 2N-r places.

Ar = A’r + A’’r = C(2N-r-1, r-1) + C(2N-r, r)

= 2N/(2N-r) C(2N-r, r). By Thm 10.1, the number of ways to seat the men is

r=0..N (-1)r (N-r)! (2N/(2N-r) ) C(2N-r, r).

p15.p15.

Ex: N=3, r=2

1

0

0

Definition (Group): A group <G, *> is a set G, together with binary operation * on G, such that

(1) * is associative, (2) There is an identity element e in G such

that e * x = x * e for all x in G, (3) For each a in G, there is an element a’ in G such that a * a’ = a’ * a = e. Definition (Group action): Let X be a set and G a

group. An action of G on X is a map * : GX X such that

(1) ex = x for all x in X, (2) (gh)(x) = g(hx) for all x in X and all g, h in

G. Under these conditions, X is a G-set.

Definition : Let X be a G-set. Let x X and g G. Define Xg= {x X : gx = x} and Gx={g G: gx = x}.

Thm: Let X be a G-set. Then Gx is a subgroup of G for each x X.

Proof: (1) e Gx, since ex=x.

(2) If g Gx, then gx = x and thus x=g-1x, which implies g-1 Gx.

(3) If g, h Gx, then gx=hx=x.

Thus (gh)(x) = g(h(x))= g(x) = x, which implies

gh Gx.

Therefore Gx is a subgroup of G.

Thm: Let X be a G-set. For a, b X, let a~b if and only if there exists g G such that ga = b. Then ~ is an equivalence relation on X.

Proof: (1) For each x X, we have ex=x, so x~x. –

reflexive. (2) If a~b, then ga = b for some g G. So b=g-1

a and b~a. –symmetric. (3) If a~b and b~c, then ga = b and hb = c for

some g, h G. So hga = hb = c and a~c. –transitive.

Def: (Orbits) Let X be a G-set. Each cell in the equivalence classes for the relation “~” is an orbit in X under G. If x X, the cell containing x is the orbit of x. Let this cell be Gx.

Thm: Let X be a G-set and let x X. Then |Gx| = (G: Gx)=|G|/| Gx|.

Proof: Want to define a 1-1 map from Gx to the left cosets of Gx. Let a Gx. Then there exists g G such that gx=a. Define the map f(a) to be the left coset gGx of Gx. If there is another h G such that hx=a, then hx=gx. So g-1hx=g-1gx, i.e., g-1hx=x, which implies g-1h Gx.

Thus h gGx and hGx = gGx. So f is well defined. Suppose a, b Gx and f(a)=f(b). Then there exist g, h G

such that gx=a and hx=b and f(a)=gGx=f(b)=hGx, which implies g=hg’ for some g’ Gx. So a=gx=hg’x=hx=b. Thus f is 1-1.

Let gGx be a left coset. Then if gx=x’, we have gGx = f (x’). Thus f maps Gx onto the collection of the left cosets of Gx.

Gx: the cell containing x. Gx={g G: gx = x}.

Thm 10.5: (Burnside) Let G be a finite group and X a finite G-set. Then the number of orbits in X under G is equal to

1/|G| g G |Xg|.Proof: Consider all pairs (g, x) where gx=x, and let N be the

number of such pairs. For each g G, it contributes |Xg| pairs. Thus N = g G

|Xg|. For each x, there are |Gx| pairs. Thus N= x X |Gx|. By the previous Thm, |Gx|=(G : Gx) =|G|/| Gx|. Then N= x X |G|/| Gx| = |G| x X 1/| Gx|. 1/| Gx| has the same value for all x in the same orbit. Thus N=|G| (number of orbits in X under G), i.e., (number of orbits in X under G) = 1/|G| N= 1/|G| g G |

Xg|.

Gx: the cell containing x.

Gx={g G: gx = x}. Xg = {x X : gx = x}