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COmbination problem

  1. May 24, 2014 #1
    Hi,
    An ice cream shop sells ten different flavors of ice cream. You order a two-
    scoop sundae. In how many ways can you choose the flavors for the sundae if
    the two scoops in the sundae are different flavors?

    That's what I did :

    9+(9-1)+(9-2)+...+(9-9)=45

    Would this be any good ?

    Thank you!
     
  2. jcsd
  3. May 24, 2014 #2

    Dick

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    It might be if you explain your reasoning. It's not the usual way to write it. Why do you think it's correct?
     
  4. May 24, 2014 #3
    Well, we need two different sorts, so we have a total combination of 9 for 10 different flavour.

    For example, if we consider each flavour a number, then : (1,2) (1,3) ... til (1,10)

    Now, if we want to see for the second flavour, we already did the combination (1,2), so we must substract 1 combination from 9.

    so we have 9+(9-1) and we continue this way.

    I hope that's what you're seeking !
     
  5. May 24, 2014 #4

    Mentallic

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    Your reasoning is solid and your answer is correct. It may also be helpful for you to know that the answer is 10 choose 2, which is denoted by [itex]^{10}C_2[/itex].
     
  6. May 24, 2014 #5

    Dick

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    Yes, that's what I'm seeking and yes, it's correct. There is a more standard way to do this. See http://en.wikipedia.org/wiki/Combination That would let you write it as 10*9/2=45. Same number.
     
  7. May 25, 2014 #6
    Ok, Thank you again. I didn't know the standard way...^^
     
  8. May 25, 2014 #7

    Simon Bridge

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    The "C" notation seems to be out of fashion, at least for starting out.
    John Allen Paulos' influence I suspect.

    The usual approach seems to be to argue there are 10 possibilities for the 1st scoop, and, since the second scoop has to be different, there are 9 remaining ... for 10x9=90 possible sundaes with 2 different flavors... however, that counts say "vanilla+ chocolate" as different from "chocolate+vanilla" ... since the order does not matter, then the required value is half that: 90/2=45.
    http://betterexplained.com/articles/easy-permutations-and-combinations/

    That was quite interesting reasoning though, I'd like to see that applied to combinations of 3 scoops.
     
  9. May 25, 2014 #8

    verty

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    This is similar to the problem of counting handshakes: if 10 people shake hands such that every person shakes hands with every other person exactly once, how many handshakes take place?
     
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