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Number of partitions of 11 elements into 3 subsets? watch

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    Hi, I am asked to compute the number of partitions of a set consisting of 11 elements into three subsets.

    I found this equation on wikipedia http://en.wikipedia.org/wiki/Stirlin...ind#Definition

    and I've tried it with 4 elements, 3 subsets and I get \frac{1}{6}(1)(1)(3^4) = 13.5 but (if i'm right) there are only 6 different partitions of 4 elements into 3 subsets:

    (12)(3)(4)

(1)(23)(4)

(1)(2)(34)

(14)(2)(3)

(13)(2)(4)

(1)(24)(3)

    (which makes more sense than 13.5 anyway since it is a whole number).


    Am I using the wrong equation? Could anyone be of help? Thank you so much
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    If you substitute into that formula correctly then you get \begin{Bmatrix} 4 \\ 3 \end{Bmatrix} = \dfrac{1}{6} \bigg( (1)(1)(3^4) + (-1)(3)(2^4) + (1)(3)(1^4) + (-1)(1)(0^4) \bigg), which correctly gives 6; I'm not sure why you stopped at the first term of the sum.
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    ahh thank you so much!!
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    (Original post by nuodai)
    If you substitute into that formula correctly then you get \begin{Bmatrix} 4 \\ 3 \end{Bmatrix} = \dfrac{1}{6} \bigg( (1)(1)(3^4) + (-1)(3)(2^4) + (1)(3)(1^4) + (-1)(1)(0^4) \bigg), which correctly gives 6; I'm not sure why you stopped at the first term of the sum.
    Sorry to bother you again I just want to make sure I've fully understood! I've just tried doing it with n=11, k=3;

    \frac{1}{6}[((1)(1)(3^{11})) + ((-1)(3)(2^{11})) + ((1)(3)(1^{11})) + ((-1)(1)(0^{11}))]

    However, being to the power of 11, I feel like this sequence should carry on? But \displaystyle \binom{3}{4} and \displaystyle \binom{3}{5} etc are impossible so I might be right in ending at \displaystyle \binom{3}{3} in the final (-1)\displaystyle \binom{3}{3}(0^{11}) = (-1)(1)(0^{11}) term?

    Thank you again!
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    (Original post by furryvision)
    Sorry to bother you again I just want to make sure I've fully understood! I've just tried doing it with n=11, k=3;

    \frac{1}{6}[((1)(1)(3^{11})) + ((-1)(3)(2^{11})) + ((1)(3)(1^{11})) + ((-1)(1)(0^{11}))]

    However, being to the power of 11, I feel like this sequence should carry on? But \displaystyle \binom{3}{4} and \displaystyle \binom{3}{5} etc are impossible so I might be right in ending at \displaystyle \binom{3}{3} in the final (-1)\displaystyle \binom{3}{3}(0^{11}) = (-1)(1)(0^{11}) term?

    Thank you again!
    Why would the series carry on? The series runs from 0 to k, and in this case k=3, so it makes sense that you have 4 terms. The fact that you have a power of 11 in there needn't have an impact on the sum (and if the 11 wasn't there then it would be worrying because it doesn't appear anywhere else!) From what I can see, you've done it right; you just have to evaluate it now.
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    (Original post by nuodai)
    Why would the series carry on? The series runs from 0 to k, and in this case k=3, so it makes sense that you have 4 terms. The fact that you have a power of 11 in there needn't have an impact on the sum (and if the 11 wasn't there then it would be worrying because it doesn't appear anywhere else!) From what I can see, you've done it right; you just have to evaluate it now.
    thank you so much very grateful!
 
 
 
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