20 Comments

  • DirectorRico

    December 11, 2013

    Was here at the beginning of the year, now I’m back for my finals. Thanks
    for this.

  • ironmantis25

    December 11, 2013

    I’m applying this to yugioh, I need a consistent deck to win.

  • Brian Veitch

    December 11, 2013

    To give some more examples, (13C3) means to grab any three ranks from the
    13 available ranks. So (13C3)(4C2)(4C1)(4C3) means grab any three ranks,
    then choose 2 of the 4 cards for the first rank, then 1 of the 4 cards for
    the second rank, and 3 of the 4 cards from the third rank. So how you set
    it up depends on what you’re hand looks like. That’s why your original
    examples were set up differently. Just to make sure, rank means a type of
    card. Ace is a rank, 2 is a rank, 3 is a rank, etc.

  • Brian Veitch

    December 11, 2013

    13C1 is just used to represent you choosing one rank (type of card). But it
    could be any rank. For example, (13C1) means it could be K, 10, Ace, 5,
    Queen, 2, etc. So (13C1)(4C1) means to choose a rank, and then one card
    from that rank. (13C1)(4C2) it means a pair from any rank. (13C1)(4C3)
    means choose a rank and then choose 3 cards from that rank. But (13C1)
    isn’t always used. For example, two pair uses (13C2)(4C2)(4C2). (13C2)
    means pick any two ranks from the 13 available ranks.

  • sadafbenaf

    December 11, 2013

    Thank you so much for the clearing, I thought I loved math! I feel like a
    child when it comes to permutation and combinations. So, 13C1 could only be
    used if we are asked for a pair, 2 pairs, 3 or four of a kind, full house,
    and royal flush?

  • Brian Veitch

    December 11, 2013

    The second example is a very specific type of 3 of a kind. It won’t have
    the (13C1) because your 3 of a kind are JACKS. Plus your 4th card is fixed
    as a Queen. This is why there is no (12C2) anymore from my example. So
    (4C1) for the queen, (4C3) for the 3 Jacks, (44C1) to find the last card
    from the remaining 44 ranks that is not a Queen or Jack. Hope this helps.
    Just remember your examples are not general 3 of a kinds. Your formulas for
    them will change because some cards are fixed.

  • Brian Veitch

    December 11, 2013

    These two examples don’t fit the general three of a kind scenario. Exactly
    three kings means the other two cards could be the same (but not Kings).
    This means you’re including some combinations that are a full house. Plus
    since your 3 of a kind are KINGS, you don’t have the (13C1) at the
    beginning. (13C1) allows for the 3 of a kind to be any card (3 Kings, 3
    10s, 3 Aces, etc).

  • sadafbenaf

    December 11, 2013

    Hi Brian, my book solves (how many different 5-hand cards are possible that
    consist of exactly 3 kings?) as (4C3)(48C2)=4512. Or, (how many different
    5-hand cards are possible that consist of 1 Queen and 3 Jacks?) as
    (4C1)(4C3)(44C1)=704 How are these different than your way of solving three
    of a kind @ 9:07

  • Mandooke Jama

    December 11, 2013

    Title says permutations and combinations but it’s just combinations

  • Brian Veitch

    December 11, 2013

    It’s the most common mistake, one I made the first time tried to do this
    problem. If you calculate it your way, you’ll get twice as many
    combinations. It’s because of the (13C1)*(12C1) part. If I remember
    correctly, this means the order of the two ranks matter. (KK994 is
    different than 99KK4) We know they are considered the same hand. It’s why
    your number is twice as much as it should be. Instead, use (13C2). This
    allows you to select 2 ranks, where order does not matter. KK994 = 99KK4

  • Peter Naugler

    December 11, 2013

    I tried the two pair example before he said the answer and I got
    13C1*4C2*12C1*4C2*11C1*4C1. Why is that wrong? Or, what do I misunderstand
    if I got that answer?

  • DirectorRico

    December 11, 2013

    Good pace. Thank you.

  • Juan Gonzalez

    December 11, 2013

    Great way to explain everything!! Your a life saver!

  • Johnny Wu

    December 11, 2013

    really help me out with my HW!

  • Dietpill

    December 11, 2013

    This video helped me out a great deal with my HW. Thank you!

  • dasnyds00

    December 11, 2013

    Thanks for the videos. They were a huge help!

  • Diana Gee

    December 11, 2013

    GREAT VIDEOS, THANK YOU!

  • David Hoeksema

    December 11, 2013

    Your 3 of a kind example is incorrect. Should be 13C1x4C3x12C1x4C1x11C1x4C1

  • Brian Veitch

    December 11, 2013

    12C1 x 4C1 x 11C1 x 4C1 = 48 x 44 is more like a permutation because if you
    switch the last two cards, this will treat it as a new hand. For example,
    the way you have it written, QQQ5J would be a different hand than QQQJ5. So
    you have to take 12C1 x 4C1 x 11C1 x 4C1 and divide by 2 to account for the
    order of the last two cards. (12C1) x (11C1) / 2 = (48 x 44) / 2 is the
    same as 12C2.

  • monalisa

    December 11, 2013

    You broke down the steps very well and I understood everything. Thank you
    so much…..

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