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Thread: Combination of Pathways to get to Result

  1. #1
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    Combination of Pathways to get to Result

    Select 4 balls at random from an infinite pool of balls containing 4 different colored balls. How many different different ways (order) you can make the selections would seem to be a combination based formula. Meaning you could select first a Blue, then a Red, then a Blue and a Yellow, but to get 2 Blues and 2 of the 3 other colors you could also pull the blues first or last for 1 pathway each (permutations).

    Answers:
    4 different colors = 24 different ways of selecting them (4*3*2*1)
    1 matching color and 2 different colors = 12 different ways of selecting them
    2 matching pairs of colored balls = 6 different ways of selecting them
    3 matching color balls and 1 other color ball = 4 ways

    What formula logically represents this problem (4,6,12,24) that would also work if you had n number of different colored balls with k number of pulls?

    Thanks in advance for your help!
    Last edited by Smix; Nov 4th 2017 at 01:34 PM.
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  2. #2
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    Re: Combination of Pathways to get to Result

    Quote Originally Posted by Smix View Post
    Select 4 balls at random from an infinite pool of balls containing 4 different colored balls. How many different different ways (order) you can make the selections would seem to be a combination based formula. Meaning you could select first a Blue, then a Red, then a Blue and a Yellow, but to get 2 Blues and 2 of the 3 other colors you could also pull the blues first or last for 1 pathway each (permutations).
    Answers:
    4 different colors = 24 different ways of selecting them (4*3*2*1)
    1 matching color and 2 different colors = 12 different ways of selecting them
    2 matching pairs of colored balls = 6 different ways of selecting them
    3 matching color balls and 1 other color ball = 4 ways
    I am in no way sure that I understand the setup here!

    All four different in $1$ way; arrange in $4!$ ways.

    Select one to match in $4$ ways, two others in $\dbinom{3}{2}$; arrange in $\dfrac{4!}{2!}$ ways.

    Select two to match in $4$ ways $\dbinom{4}{2}$, two others in $\dbinom{3}{2}$; arrange in $\dfrac{4!}{2!\cdot 2!}$ ways.

    Select one to match times in $4$ ways ,one others in $3$ ways; arrange in $\dfrac{4!}{3!}$ ways.

    Select one to match four times in $4$ ways; arrange in $1$ way.

    As to express that in one formula, I have no clue.
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  3. #3
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    Re: Combination of Pathways to get to Result

    Quote Originally Posted by Plato View Post
    I am in no way sure that I understand the setup here!

    All four different in $1$ way; arrange in $4!$ ways.

    Select one to match in $4$ ways, two others in $\dbinom{3}{2}$; arrange in $\dfrac{4!}{2!}$ ways.

    Select two to match in $4$ ways $\dbinom{4}{2}$, two others in $\dbinom{3}{2}$; arrange in $\dfrac{4!}{2!\cdot 2!}$ ways.

    Select one to match times in $4$ ways ,one others in $3$ ways; arrange in $\dfrac{4!}{3!}$ ways.

    Select one to match four times in $4$ ways; arrange in $1$ way.

    As to express that in one formula, I have no clue.
    To express them in one formula, use the sum principle:

    $\dbinom{4}{4}4!+\dbinom{4}{3}\dbinom{3}{2}\dfrac{ 4!}{2!}+\dbinom{4}{2}\dfrac{4!}{2!2!}+\dbinom{4}{2 }\dbinom{2}{1}\dfrac{4!}{3!}+\dbinom{4}{1}\dfrac{4 !}{4!}=4^4$

    That's the number of ways to choose 4 distinct balls plus the number of ways to choose 3 distinct balls where one color is repeated. Plus the number of ways to select 2 distinct colors with each repeated plus the number of ways to select two colors with only one color repeated plus the number of ways to select 4 balls of the same color.
    Last edited by SlipEternal; Nov 4th 2017 at 08:27 PM.
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    Re: Combination of Pathways to get to Result

    Quote Originally Posted by SlipEternal View Post
    To express them in one formula, use the sum principle:

    $\dbinom{4}{4}4!+\dbinom{4}{3}\dbinom{3}{2}\dfrac{ 4!}{2!}+\dbinom{4}{2}\dfrac{4!}{2!2!}+\dbinom{4}{2 }\dbinom{2}{1}\dfrac{4!}{3!}+\dbinom{4}{1}\dfrac{4 !}{4!}=4^4$

    That's the number of ways to choose 4 distinct balls plus the number of ways to choose 3 distinct balls where one color is repeated. Plus the number of ways to select 2 distinct colors with each repeated plus the number of ways to select two colors with only one color repeated plus the number of ways to select 4 balls of the same color.
    I think I knew that, because Smix has described all possible configurations of 4-strings from four.
    I was under the impression Smix wanted a closed expression(function) into which a variable is put and the number of ways is the output.

    Perhaps I am reading too much into it?
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  5. #5
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    Re: Combination of Pathways to get to Result

    Quote Originally Posted by Plato View Post
    I think I knew that, because Smix has described all possible configurations of 4-strings from four.
    I was under the impression Smix wanted a closed expression(function) into which a variable is put and the number of ways is the output.

    Perhaps I am reading too much into it?
    Oh, yeah. If that's what he wants I have no clue
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    Re: Combination of Pathways to get to Result

    Yeah I was hopeful for a N, K (and likely J) expression that would be applicable to pulling any number of permutations of balls from various different color sets. I know the answers in the simple cases.
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  7. #7
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    Re: Combination of Pathways to get to Result

    Say you have $n $ different colors, and you want to pick $k$ balls from your infinite supply. With $1\le r \le n$, any integral solution to the Diophantine equation $\displaystyle \sum_{r=1}^n x_r =k $ with nonnegative numbers represents an arrangement of the number of balls of each color you want. The number of ways to choose $k $ balls with that arrangement of colors is given by:
    $\dbinom{n}{x_1,x_2,\ldots , x_n, n-k} $
    (This is a multinomial coefficient)
    The number of ways to arrange them is given by $\dbinom {k}{x_1,\ldots, x_n} $.
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