Inclusion and exclusion principle; inversion formulae Thm Let S be an N-set; E i ’s, i=1..r, be subsets of S. For any subset M of [r], define N(M) = |  i  M E i | and for j=0..r, define N j =  |M|=j N(M). Then the number of elements of S not in any E i is N-N 1 +N 2 -N 3 +…+(-1) r N r. (*) Pf: If x is in S and not in any E i, then x contributes 1 to (*). If x is in S and in exactly k of E i, then the contribution to (*) is

p2. Eg Let d n denote the number of permutations  of 1, 2,…, n such that  (i)  i. Let E i be the subset of those permutations  with  (i)=i. By Thm 10.1 we have:

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?

p4.

p5.

Eg. (The Euler function)

p7.

p8.

 d | n  d’ | d f(d, d’) =  d’ | n  d | n/d’ f(dd’, d’) Example: Take n= d d’

Example: Count the number of N n 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 N n =  d|n M(d). Why? Observe that 2 n =  d|n dM(d). By Thm 10.4, nM(n) =  d|n μ(d) 2 n/d. Why? So, M(d) = (1/d)  d’|d μ(d’) 2 d/d’. Thus, N n =  d|n M(d) =  d|n M(d) =  d|n (1/d)  d’|d μ(d’) 2 d/d’ =  d|n (1/d)  d’|d μ(d/d’) 2 d’ =  d’|n  d|n/d’ (1/dd’) μ(d) 2 d’ =  d’|n 2 d’ /d’  d| n/d’ μ(d)/d =  d’|n 2 d’ /d’  (n/d’)/(n/d’) = (1/n)  d’|n 2 d’  (n/d’)

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 A r 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 A r = 2 r (2N-r-1)!. By Thm 10.1, there are  r=0..N (-1) r C(N, r) A r 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 A r 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. A r 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 i-1 2i 0 1 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. A r = 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. Ex: N=3, r=

Definition (Group): A group 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 * : G  X  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 X g = {x  X : gx = x} and G x ={g  G: gx = x}. Thm: Let X be a G-set. Then G x is a subgroup of G for each x  X. Proof: (1) e  G x, since ex=x. (2) If g  G x, then gx = x and thus x=g -1 x, which implies g -1  G x. (3) If g, h  G x, then gx=hx=x. Thus (gh)(x) = g(h(x))= g(x) = x, which implies gh  G x. Therefore G x 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: G x )=|G|/| G x |. Proof: Want to define a 1-1 map from Gx to the left cosets of G x. Let a  Gx. Then there exists g  G such that gx=a. Define the map f(a) to be the left coset gG x of G x. If there is another h  G such that hx=a, then hx=gx. So g -1 hx=g -1 gx, i.e., g -1 hx=x, which implies g -1 h  G x. Thus h  gG x and hG x = gG x. 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)=gG x =f(b)=hG x, which implies g=hg’ for some g’  G x. So a=gx=hg’x=hx=b. Thus f is 1-1. Let gG x be a left coset. Then if gx=x’, we have gG x = f (x’). Thus f maps Gx onto the collection of the left cosets of G x. Gx: the cell containing x. G x ={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 |X g |. Proof: Consider all pairs (g, x) where gx=x, and let N be the number of such pairs. For each g  G, it contributes |X g | pairs. Thus N =  g  G |X g |. For each x, there are |G x | pairs. Thus N=  x  X |G x |. By the previous Thm, |Gx|=(G : G x ) =|G|/| G x |. 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 |X g |. Gx: the cell containing x. G x ={g  G: gx = x}. X g = {x  X : gx = x}