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Math Help - binominal coefficients

  1. #1
    baz
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    binominal coefficients

    Need help on this one,

    Let
    A be a finite set with cardinality |A| = n. Prove that A has exactly
    C(n,k) subsets ofcardinality k. ( n and k are arbitrary integers such that
    0 =< k = < n)

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    Quote Originally Posted by baz View Post
    Need help on this one,

    Let A be a finite set with cardinality |A| = n. Prove that A has exactly
    C(n,k) subsets ofcardinality k. ( n and k are arbitrary integers such that
    0 =< k = < n)


    This is straightforward induction on |A| + some basic identities of binomial coefficients ...where are you stuck?

    Tonio
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  3. #3
    baz
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    it means I know to little about binominal coefficients to answer that.
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    Quote Originally Posted by baz View Post
    it means I know to little about binominal coefficients to answer that.
    If that statement is in fact the case, then why in the world are you trying to do these problems?
    Moreover, given the fact that we do not offer tutorials, what do you expect from us.
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  5. #5
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    Quote Originally Posted by Plato View Post
    If that statement is in fact the case, then why in the world are you trying to do these problems?
    Moreover, given the fact that we do not offer tutorials, what do you expect from us.
    I dont know how you can say that 'PLATO'. It does not mean that you stop trying if you are struggling with something.
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    Quote Originally Posted by baz View Post
    Need help on this one,

    Let
    A be a finite set with cardinality |A| = n. Prove that A has exactly
    C(n,k) subsets ofcardinality k. ( n and k are arbitrary integers such that
    0 =< k = < n)

    Hi baz,

    If we take k elements from n in all possible arrangements of the k elements,
    there will be

    n(n-1)(n-2)(n-3)........(n-[k-1]) many arrangements.

    as there are k factors in the above expression,
    since by arranging them,
    there are n choices for the 1st element,
    having chosen one there are (n-1) remaining choices for the 2nd,
    (n-2) remaining choices for the 3rd....etc

    Since, we are not multiplying by (n-k)((n-[k+1])(n-[k+2]).....3(2)1

    we can write that as

    \frac{n!}{(n-k)!}

    This has counted all possible groups of k distinct elements from the n
    in all possible orders


    hence we need to unarrange them, to find the number of subsets
    with k elements.
    Therefore we divide by k!

    \frac{n!}{(n-k)!k!} is written \binom{n}{k} or C(n,k) or Nc_k

    The number of ways to choose k elements from a total of n is \binom{n}{k}
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    Quote Originally Posted by baz View Post
    I dont know how you can say that 'PLATO'. It does not mean that you stop trying if you are struggling with something.

    Of course not, but then you first must make sure you know all the required basic stuff to at least understand and begin to attack the problem...and if you don't know/understand the basics then you first must study them

    Tonio
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