clarity_ | given 52 cards with 13 values and 4 suits... how many possible sets of 3 cards of the same suit are there |

majd | TRWBW, ok, thanks |

clarity_ | is it 4*13*12*11 ? |

TRWBW | np clarity_: 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 whatever clarity_: but i think you got it |

mysteryx | does anyone here know rf theory...? |

clarity_ | 26*25*24 I believe 26 possiblities then 25 then 24 but 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! permutations so dividing by 3! would give you the one set right? |

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 think okay... 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 = 39 correct? |

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)? |