Combinatorial Mathematics III B.Sc(Maths)

Inclusion and exclusion principle Let A be cthe collections of objects given along with a list of r properties which the objects may or may not posses. Find the number of objects which posses atleast one of properties. Proof: Let N(i,j,…r) denote the number of objects which posses each of i,j,,,,,,,r properties. The number of objects possessing atleast one of properties=N(1)+N(2)+.N(r){N(1,2)+N(2,3)+….. N(r-1,r)}+{N(1,2,3)+N(2,3,4)+……N(r-2,r- 1,r)}+……+(-1) r-1 N(1,2,…….r) (1)

If an object possess none of the properties it is clear that it contributes nothing to (1) If an object possess t>1 properties then it contributes 1 to (1).For its contribution is =[t- -……] = - [-t+ +…. =1-[1- [-t+ +….]]=1-(1-1) t =1 Problems How many integers from 1 to 1000 are divisible by none 3,7,11……

Solution: Let N(R) denote the number of numbers between 1 and 1000 which are divisible by r. N(3)=333, N(7)= 142, N(11)=90 N(3,7)=N(21)=47, N(7,11)=N(77)=12, N(11,3)=N(33)=30 and N(3,7,11)=N(231)=4 N(3 or 7 or 11)=N(3)+N(7)+N(11)-N(3,7)- N(7,11)-N(11,3)+N(3,7,11) = =480 Number of numbers divisible by none of 3,7,11= =520.

2. How many permutations are there of the digits 1,2,…8 in which none of the patterns12,34,56,78 appears. Solution: Take (12) as single unit and other 6 numbers as 6 other units. These 7 units can be permuted in 7! Ways. N(12)=N(34)=N(56)=N(78)=7! N(12,34)=N(34,56)=N(56,78)=N(78,12)=6! N(12,56)=N(34,78)=6! N(12,34,56)=N(34,56,78)=N(12,34,78)=5! N(12,56,78)=5! and N(12,34,56,78)=4!

N(12 or 34 or 56 or 78)= [N(12)+N(34)+N(56)+N(78)]- [N(12,34)+..]+[N(12,34,56)+….]-[N(12,34,56,78)] = Number of ways in which none of given pattern occurs =total number of permutation = Prove that number of derangements of n symbols = n![1- +]

Solution: Total number of permutation of n symbols = n! Take n! permutation as n objects. An object has i th property with i th symbols remains in i th place N(i)=number of objects having i th property=(n-1)! This type occurs in nc1 ways. Similarly N(i,j)=number of objects having properties i&j =(n-2)! This type occurs in nc2 ways. N(i,j,k) =number of objects having properties i,j&k =(n-3)!

This type occurs in nc3 ways and so on. Number of objects having atleast one of properties =N(1 or 2 or…n) =N(1)+N(2)+………N(n)- [N(1,2)+N(2,3)+………N(n-1,n)]+…..+ (-1) n N(1,2….n)=(n-1)! -(n-2)! +… =n!- + -……..