Recent content by guropalica

C(n,k), n-k the number of elements of Y, 2^(n-k) the number of subsets...by binomial I got 3^n! Thanks a lot guys

C(n,k), n-k the number of elements of Y, 2^(n-k) the number of subsets...by binomial I got 3^n! Thanks a lot guys

n-1 a typo...There are k^2 possibilities for X, and C of k from n for Y I think! PS.Can we go more direct with hints I'm deadline's due 02:15 GMT :)

If x has k elements, than Y has n-k to n-k elements at most! Minimum value for k is 1, and max is n-1 It's pretty much pointless to write about 2 elements since that way I'd have only 2 cartesian products only since by the condition sets are not subsets so they can't be equal to the parent ie...

Set S with n elements! Set size of {<x,y> | (X,Y are proper subsets of S), (X union Y = S)! I tried doing something, but I'm stuck staring at a closed door, so I need a fresh start! Any hints would be appreciated!

Got it now :) let's denote the initial statement as K + L, so we have sth like 24l + 7k = 7(k+l) + 17k, sum of two numbers divisible by 17 is divisible by 17, anyway thx

I proved the base case n=1, and then I try doing the step case assuming that it satisfies for any k, then I try proving it by k+1. I got sth like 24 * (3 * 5^(2k+1)) + 7 * 2^(3k +1) Now I'm stuck can't continue, I don't have any ideas :/ btw the initial equation is 3 * 5^(2n+1) + 2^(3n+1) !

Proof by induction that 3 * 5^2n+1) + 2^3n+1 is divisible by 17! Thanks in advance guys