Ordered and Unordered Sets, in simple words, can be seen as follows :
Ordered sets :
If for a given situation,a set of n elements satisfies the problem solution then all permutations of those n elements that satisfy the problem solution are to be counted as different i.e
a,b is treated different than b,a ; a,b,c is counted different from a,c,b or b,a,c or b,c,a or c,a,b or c,b,a
Unordered sets :
If for a given situation,a set of n elements satisfies the problem solution then any permutation of that set satisfying the problem solution should be considered as the same and not counted again.
a,b is same as b,a and a,b,c is same as other (3! - 1) = 5 permutations.
POINT 1)
Number N as a product of its two factors.
f : Total number of factors of N.
N = a*b
N is a perfect square :
If N is a perfect square it contains one solution as N = a*a = a^2 and rest are (a,b) distinct format.
Hence,
Ordered(distinct) = (f - 1)
Ordered(not necessarily distinct) = f
Unordered(distinct) = (f - 1)/2
Unordered(not necessarily distinct) = (f - 1)/2 + 1 OR (f + 1)/2
N is not a perfect square
Ordered(any case) = f
Unordered(any case) = f/2
example :
N = 100 = 2^2 * 5^2
f = 3*3 = 9
100 = a*b
Then if a is some factor of 100,then b = 100/(that factor) and vice versa.
Hence we have as many (a,b) sets satisfying the equation as are the number of factors of N.
But since N = 100 = perfect square hence it has one solution : 10,10 counted twice.
1*100,100*1,2*50,50*2,...10(first)*10(second),10(second)*10(first)
So,
Ordered(distinct) = f - 1 = 8
Ordered(not necessarily distinct) = f = 9
Unordered(distinct) = (f - 1)/2 = 4
Unordered(not necessarily distinct) = (f - 1)/2 + 1 OR (f + 1)/2 = 5
POINT 2)
Number of ways to resolve N into product of two co-primes = 2^(n - 1) where n is the number of distinct primes in the prime factorization of N.
example :
N = 12 = 2^2 * 3
Then, answer = 2^(2 - 1) = 2 ; As there are two primes : 2 and 3 in factorization of N
We can see that as :
1*12 : Co-prime
2*6 : HCF(2,6) = 2
3*4 : Co-prime
2 sets.
POINT 3)
Number of ways to resolve N into product of three co primes.
example :
N = 2^3 * 3^2 * 5^4
N = abc such that HCF(a,b,c) = 1 i.e they are co-prime.
a = 2^x1 * 3^y1 * 5^z1
b = 2^x2 * 3^y2 * 5^z2
c = 2^x3 * 3^y3 * 5^z3
x1 + y1 + z1 = 3 => 5C2 ways. Out of this when none of them is zero needs to be removed as in that case
HCF would not be 1.
So, 5C2 - 1 = 9 ways
Similarly,
x2 + y2 + z2 = 2 => 4C2 = 6 ways. With none of them zero the equation cannot hold for non negative integers.
Hence nothing is to be subtracted.
Also,
x3 + y3 + z3 = 4 => 6C2 - 3 = 12 ways
Ordered = 9*6*12 = 648
Unordered = (648 - 3!/2!*4)/3! + 4 = 110
[1,1,n ; 3,3,x ; 25,25,y ; 75,75,z ] being the 4 triplets to be removed for unordering ]
That brings us to close of the 3 part series of Factors - Fundamentals article. These 3 part should allow you to solve most of the problems related to Factors of a number barring a few typical concepts which will be discussed as separate special article devoted to that topic alone later down the line. Till then, keep following the blog for such illustrative concepts.
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AS
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