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Combinatorics discrete math

  1. Mar 7, 2014 #1
    Hi all I need some assistance
    1. The problem statement, all variables and given/known data with the attempt

    How many 5-digit briefcase combinations contain

    1. Two pairs of distinct digits and 1 other distinct digit. (e.g 12215)
    I wasn't sure on which approach was correct.

    10 * 9 * 8 (because there are three distinct digits)

    or

    10C2 * 5C2* 3C2 * 8 (because you have to take into account how the doubles can be orientated)

    2. A pair and three other distinct digits. (e.g 27421)
    same issue as above

    10*9*8*7

    or

    (5C2*10) * 9 * 8 * 7

    lastly
    I have found this question before but I couldn't get an explanation:

    How many ways are there to pick a collection of 12 coins from piles of pennies, nickels, dimes, quarters, and half-dollars? Base on the following condition:

    1. there are only 10 coins in each pile.
    16C4 - 5^2 because it's the total minus how many ways I can get from the 11th coin and the 12th coin. = 1795
    2. There are only 10 coins in each pile and the pick must have at least one penny and two nickels?
    1795 - 13C4(?)
    my logic is that it's because 12-3+4 C 4 but I'm not sure if I have to set it to 12 or 10.

    Thank you!
     
  2. jcsd
  3. Mar 7, 2014 #2
    I only have a minute, but I'll see what I can get through:

    1. First let's pick which digits are going to be used. There are 10 choices for the first paired digits, 9 choices for the second paired digits, and 8 choices for the lone digit. So there are ##10*9*8## ways to choose these digits. Now we have to order them. If we could distinguish our five numbers, the number of ways to order them would be ##5!##. But we have those two indistinguishable pairs. So, for example, putting our first pair in the second and fourth position, we be counting this twice in the ##5!##. So we need to divide by two for each of the pairs. Thus the number of ways to order our 5 numbers accounting for the fact that there are two pairs is: ##\frac{5!}{2*2}##.

    Thus the total number of combinations is: ##10*9*8*\frac{5!}{2*2}##.
     
  4. Mar 7, 2014 #3
    for the positioning

    would it matter if the doubles were not next to each other for ex.

    12125 or 13212?
     
  5. Mar 7, 2014 #4

    haruspex

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    That's not quite right. E.g. 11223 gets counted both by picking two 1s, then two 2s and by picking them in the other order.
     
  6. Mar 8, 2014 #5
    AH I solved it!, Thank you for replying to the posts!
     
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