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Math Help - Basic probability problem

  1. #1
    Junior Member 22upon7's Avatar
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    Basic probability problem

    Hi,

    I'd really like some help on how to work this sort of sums out. A step by step explanation would be great!

    Allen is making a bead neclace using four different coloured beads. How many different ways can he arrange these four beads on the necklace?

    I think the answer is 16, but I'm not sure. I worked it out by listing every possible combination - but I think there might be a formula to solve questions like this.

    Thanks!

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  2. #2
    Member
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    Sacramento, CA
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    This is a combinatorial problem (permutation, specifically), common in building an intuition with probabilities. There are formulas for how to calculate these (permutation of n terms is n!, i.e., n factorial), but with small examples like this it is always good to work it out and see the result first-hand. Let A, B, C, and D stand for the four possible beads. The possible (4! = 24) combinations are then determined by the possible position slots that can be filled. We will have 4 "bases" as I think of them. There is an A base, B base, C base, and D base. In each of these bases are 6 possible combinations of the remaining three slots. Thus, we have another calculation 4! = 4*6 = 24. Once we have a base, we define three other slots. So, for example, consider the A base. Once A is in the first slot, the other three can be filled with a B second, C second, and D second. In each of these that only leaves two remaining. We can actually draw this process like a tree: see here.

    I will list the terms including their base and 2nd slots.

    Spoiler:
    ABCD (A base + B-2nd + remainder)
    ABDC (A base + B-2nd + inverted remainder)
    ACBD (... C-2nd + remainder)
    ACDB (... C-2nd + inverted remainder)
    ADBC (D-2nd)
    ADCB (D-2nd inverted)


    As you can see, there are six total in the A base. We get the same result with the others.

    Spoiler:
    BACD CABD DABC
    BADC CADB DACB
    BCAD CBAD DBAC
    BCDA CBDA DBCA
    BDAC CDAB DCAB
    BDCA CDBA DCBA


    So as you can see, setting them up is just a simple computation of filling slots and ordering what pieces you have left after you made a choice for a slot. Doing it in a uniform manner like I did above (listing them in alphabetical order, if you didn't notice), helps when it is possible. Now consider other small examples like a 3-slot problem (a subset of this one, actually). Except now think of it in terms of numbers. How many ways can you order the sequence (0, 1, 2)? Clearly it will be smaller than 4!. In fact, we already know the answer, but why is it 3!? Well, think about slots again. There's going to be 3 bases in this permutation, and that only leaves two remainders which can be listed in order or inverted. Thus, for each base we only really have is two choices. Thus, 3 * 2 = 3! = 6.

    Spoiler:
    012
    021
    102
    120
    201
    210
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  3. #3
    Junior Member 22upon7's Avatar
    Joined
    Nov 2008
    Posts
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    Quote Originally Posted by bryangoodrich View Post
    This is a combinatorial problem (permutation, specifically), common in building an intuition with probabilities. There are formulas for how to calculate these (permutation of n terms is n!, i.e., n factorial), but with small examples like this it is always good to work it out and see the result first-hand. Let A, B, C, and D stand for the four possible beads. The possible (4! = 24) combinations are then determined by the possible position slots that can be filled. We will have 4 "bases" as I think of them. There is an A base, B base, C base, and D base. In each of these bases are 6 possible combinations of the remaining three slots. Thus, we have another calculation 4! = 4*6 = 24. Once we have a base, we define three other slots. So, for example, consider the A base. Once A is in the first slot, the other three can be filled with a B second, C second, and D second. In each of these that only leaves two remaining. We can actually draw this process like a tree: see here.

    I will list the terms including their base and 2nd slots.

    Spoiler:
    ABCD (A base + B-2nd + remainder)
    ABDC (A base + B-2nd + inverted remainder)
    ACBD (... C-2nd + remainder)
    ACDB (... C-2nd + inverted remainder)
    ADBC (D-2nd)
    ADCB (D-2nd inverted)


    As you can see, there are six total in the A base. We get the same result with the others.

    Spoiler:
    BACD CABD DABC
    BADC CADB DACB
    BCAD CBAD DBAC
    BCDA CBDA DBCA
    BDAC CDAB DCAB
    BDCA CDBA DCBA


    So as you can see, setting them up is just a simple computation of filling slots and ordering what pieces you have left after you made a choice for a slot. Doing it in a uniform manner like I did above (listing them in alphabetical order, if you didn't notice), helps when it is possible. Now consider other small examples like a 3-slot problem (a subset of this one, actually). Except now think of it in terms of numbers. How many ways can you order the sequence (0, 1, 2)? Clearly it will be smaller than 4!. In fact, we already know the answer, but why is it 3!? Well, think about slots again. There's going to be 3 bases in this permutation, and that only leaves two remainders which can be listed in order or inverted. Thus, for each base we only really have is two choices. Thus, 3 * 2 = 3! = 6.

    Spoiler:
    012
    021
    102
    120
    201
    210
    Thanks so much! That was very helpful
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