## #math - Mon 12 Mar 2007 between 01:34 and 01:52

### NY Lost Funds

 clarity_ given 52 cards with 13 values and 4 suits... how many possible setsof 3 cards of the same suit are there majd TRWBW, ok, thanks clarity_ is it 4*13*12*11 ? TRWBW npclarity_: not unless you count the order of the cards clarity_ since there's 13*12*11 different ways to pull 3 cards of the same suit... then there's 4 different suits? TRWBW clarity_: set means no order, so you need to divide by 3! bung TRWBW, ok i think i can grasp the function you've written, is that something general for halfturns? majd aaaaah i get itt...so partial derivative is the derivative parallel to a plane, gradient is the derivative in any direction... clarity_ so 13*12*11/3*2 ?them times 4?i'm not sure where the 3! is coming from?do you care to explain TRWBW majd: the neat think is that a vector + dot product captures it. that's what makes it all worthwhile. you might guess that you have to differentiate in any possible direction, but it turns out if you do it in just 3 directions, you capture all the others.clarity_: forget suits, how many ways can you pick a set of 3 letters? clarity_ ah... n choose k = n!/k!(n - k)!3 letters given 13? TRWBW clarity_: sure or 26 whateverclarity_: but i think you got it mysteryx does anyone here know rf theory...? clarity_ 26*25*24 I believe26 possiblities then 25 then 24but that's including order TRWBW clarity_: nope, that's words, not sets. setwise {'c','a','t'} is the same as {'a','c','t'} majd TRWBW, yeah that's pretty cool TRWBW clarity_: that's where the 3! come from clarity_ ah!so you're counting all the permutations of the set... so if it's 3 elements you're counting 3! permutationsso dividing by 3! would give you the one setright? TRWBW clarity_: yeah, it's 4*13*12*11/3!clarity_: if it was 6 cards, it would be 4*13*12*11*10*9*8/6! clarity_ awesome! I've been having a lot of trouble trying to figure out why there's k! and it makes sense now...i thinkokay... 52 cards 13 values 4 suits. sets of 3 cards with the first 2 suits different from the 3rd is: 4*(13*12*39/3!) since 52 - 13 = 39correct? TRWBW clarity_: pick the 3rd first. clarity_ hrm 4*((13*12/2!)*(39/1!)) cld for a condition like h(x,y,z) = 0 where y and z are functions of x, what do i need in order to make claims about the uniqueness of solutions z(x), given the function f(x)?