Recent content by JessBrown

So X, Y = non-empty sets, and f = function (X,Y) I have to show that f(A and B) (c underlined) f(A) and f(B) using element argument. I have no idea - started with choosing x as a particular but arbitrarily chosen element of f(A and B) so now I have to show it is in f(A) and f(B)...

Thank you so much for your help! I kinda get it now! Thanks!

Nope, still no idea... i know that's the binomial theorem... don't quite get how that helps with the m.2^(m-1) bit...

Sooooo I wrote them all down... and the answer I got was still 216, as there are nine combinations for each of the first row's orders... so 24 ways to order the first row, means 24x9 ways to order in total... the first way was so much quicker though! eg. 1234 2143 2341 2413 3142 3412...

Ok so if I go a random set of numbres, [7654] by applying the first permutation 'a = (45)(67)', this becomes [6745] then apply the second permutation 'b=(46)(57)', it becomes [4567] which is equal to taking the first set of numbers [7654] and applying 'c=(47)(56)' to get [4567]. Is this...

Homework Statement For each m (greater than or equal to) 1, show that Sum (from k=0 to m) k . (m choose k) = m . 2^(m-1) Homework Equations The Attempt at a Solution I have tried to solve this through induction, proving for m=1, and then showing the statement for m=n, then trying...

Thank you for replying and clearing that up, I still don't entirely understand (I'm so sorry, I'm really trying to!) So if I'm combining (45)(67) and (46)(57) to get (47)(56) I go (45)(67) (the original) becomes (54)(76) then combined with (46)(57) becomes (76)(54) which then I am stuck...

So there are four books to be read by 2 people, same time to read a book each, and the question is how many different ways can they read the four books? So I firstly took the approach that you have: Person 1: Spot1 Spot2 Spot3 Spot4 Person 2: Spot1 Spot2 Spot3 Spot4 And then filling in...

Thanks, I think its just the wording and I'm interpreting it wrong, cos it says show how when (45)(67) is multiplied by (46)(57) the result is (47)(56) So to write it out I say (45)(67) goes to (54)(76) then when multiplied by (46)(57) this becomes (76)(54) which then when apply c...

Thanks, I'm having trouble with another question though, when it says that if I have three different permutations, a = (45)(67) b = (46)(57) c = (47)(56) I have to multiply them to show that if multiply any two, you'll get the third... Taking what you've said above, 4567 x (45)(67)...

Hi I'm having a little bit of trouble understanding about permutations, and how you multiply them? Say I have the permutations a = (45)(67), b = (46)(57) how do I multiply them? Thanks heaps! Jess