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Math Help - (n choose k) How can I work this out?

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    (n choose k) How can I work this out?



    I know what n choose k means but usually I get given both the values.
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    re: (n choose k) How can I work this out?

    Quote Originally Posted by uperkurk View Post

    I know what n choose k means but usually I get given both the values.
    This means 8!(N-8)!=4!(N-4)!

    From that we get, 8\cdot 7\cdot 6\cdot 5=(N-4)(N-5)(N-6)(N-7)
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    re: (n choose k) How can I work this out?

    I don't get it... what value is N?

    But how do I compute that using my calculator? N still does not have a value, you're saying 8x7x6x5 = (N - 4) but what is N?
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    re: (n choose k) How can I work this out?

    Quote Originally Posted by uperkurk View Post


    I know what n choose k means <--- ähemm?
    but usually I get given both the values.
    Quote Originally Posted by uperkurk View Post
    I don't get it... what value is N?

    But how do I compute that using my calculator? <--- you don't have to use a calculator!
    N still does not have a value, you're saying 8x7x6x5 = (N - 4) but what is N?
    1. Definbition: \displaystyle{{n \choose k} = \frac{n!}{k! \cdot (n-k)!}}

    With your equation you get:

    \displaystyle{\frac{n!}{8! \cdot (n-8)!} = \frac{n!}{4! \cdot (n-4)!}}

    2. Divide through by n!. Cross-multiply and you'll get:

    \displaystyle{\frac{(n-4)!}{(n-8)!} = \frac{8!}{4!}~\implies~\frac{(n-4) \cdot (n-5) \cdot (n-6) \cdot (n-7) \cdot (n-8) \cdot (n-9) \cdot ... \cdot 1}{(n-8) \cdot (n-9) \cdot ... \cdot 1} = \frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot ... \cdot 1}{4 \cdot 3 \cdot ... \cdot 1}}

    3. Cancel equal factors and you'll get:

    \displaystyle{(n-4) \cdot (n-5) \cdot (n-6) \cdot (n-7) = 8 \cdot 7 \cdot 6 \cdot 5

    and that's the equation Plato has posted in his post (and that's what you should have got because you know what n choose k means)

    4. This is a quartic equation in n. By first inspection you get n = -1 as a solution. But since \displaystyle{n \in \mathbb{N}} this solution isn't valid.
    You'll get the solution by comparison of the factors. And that's your turn.
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    re: (n choose k) How can I work this out?

    I've never used that formular when solving these equations. Take for example 10 choose 7. I just do 10! / 7! / 3! and I get 120
    and for example 20 choose 5 = 20! / 5! / 15! = 15,504

    I don't know how to use that formular to answer these questions :/
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    re: (n choose k) How can I work this out?

    Quote Originally Posted by uperkurk View Post
    I've never used that formular when solving these equations. Take for example 10 choose 7. I just do 10! / 7! / 3! and I get 120
    and for example 20 choose 5 = 20! / 5! / 15! = 15,504 <--- you use the equal sign as a command to evaluate a constant term, but that's not solving an equation for a variable

    I don't know how to use that formular to answer these questions :/
    There isn't much of a difference between the formula you use and mine:

    \displaystyle{{n \choose k} = \frac{n!}{k! \cdot (n-k)!}}

    with the values of your 1st example:

    \displaystyle{{10 \choose 7} = \frac{10!}{7! \cdot (10-7)!}} = (10! / 7!) / 3!

    Instead of using fixed values your equation (from your 1st post) contains a variable n. So you have to transform the given equation (which is only possible using the general form) such that you are able to determine a value for n. That's what I've done for you. Now go ahead and solve for n.
    Last edited by earboth; May 14th 2012 at 02:20 AM.
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    re: (n choose k) How can I work this out?

    Quote Originally Posted by uperkurk View Post
    I've never used that formular when solving these equations. Take for example 10 choose 7. I just do 10! / 7! / 3! and I get 120 and for example 20 choose 5 = 20! / 5! / 15! = 15,504
    I don't know how to use that formular to answer these questions :/
    The reason that I answered the way I did is that I suspected that you may not have understood the principle.
    Actually this is a vary simple question.
    In \binom{N}{9}=\binom{N}{6}, we look at it and see at once N=15.
    To understand why, one has to understand how \binom{N}{k}=\frac{N!}{k!(N-k)!} works.

    It is easy to prove that \binom{N}{k}=\binom{N}{N-k}

    So if \binom{N}{k}=\binom{N}{j} then k=N-j or N=k+j.
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    re: (n choose k) How can I work this out?

    Quote Originally Posted by Plato View Post
    It is easy to prove that \binom{N}{k}=\binom{N}{N-k}

    So if \binom{N}{k}=\binom{N}{j} then k=N-j or N=k+j.
    Strictly speaking, \binom{N}{k}=\binom{N}{N-k} is not sufficient to conclude that \binom{N}{k}=\binom{N}{j} implies N=k+j. We could also use the fact that for a given N, the sequence \binom{N}{n} for n = 0, ..., N goes first up and then down, so that for each M there exist at most two n's such that \binom{N}{n}=M.

    Quote Originally Posted by Plato View Post
    From that we get, 8\cdot 7\cdot 6\cdot 5=(N-4)(N-5)(N-6)(N-7)
    Quote Originally Posted by uperkurk View Post
    I don't get it... what value is N?

    But how do I compute that using my calculator? N still does not have a value, you're saying 8x7x6x5 = (N - 4) but what is N?
    To repeat, 8\cdot 7\cdot 6\cdot 5=(N-4)(N-5)(N-6)(N-7) is not an equation of the form N = ..., which allows calculating N immediately. Rather, it is a constraint on N (of the same sort as N^2 + 3N - 4 = 0 or 2N + 3 = 5) and has to be solved first. It is an equation of fourth degree, and two solutions can be easily guessed (earboth noted that N = -1 is a solutions). The two other solutions turn out to be complex numbers, so they are irrelevant for the problem. However, Plato's method at the top of the post is easier.
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    re: (n choose k) How can I work this out?

    Ok so here I see how you have come to 15, just add the two numbers together, I'm not all that familiar with mathematic notation, what does the . mean between 8.7.6.5 and the thing that throws me off is you're saying N - 5 or whatever

    So I can put this into the formular which would look like this

     \binom{10}{7}=\frac{10!}{7!(10-7)!}


    But if I just said

    \binom{N}{7}=\binom{N}{9}

    I don't understand your formular, I'm really bad at reading formulars


    P.S That formular still does not make sense to me... in my calculator I type, 10! divided by 7! divided by 10 - 7!

    I don't think I'm reading the formular correctly
    Last edited by uperkurk; May 14th 2012 at 06:56 AM.
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    re: (n choose k) How can I work this out?

    Quote Originally Posted by uperkurk View Post
    But if I just said

    \binom{N}{7}=\binom{N}{9}
    The answer is N=7+9=16
    Throw your calculator away or at least put in a drawer.
    You don't need it for these problems.
    You have a brain.
    Thanks from mash
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    re: (n choose k) How can I work this out?

    What so literally that's all you do? Is add the two k numbers and you have your N number?

    \binom{N}{3}=\binom{N}{7}

    N = 10?
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    re: (n choose k) How can I work this out?

    Quote Originally Posted by uperkurk View Post
    What so literally that's all you do? Is add the two k numbers and you have your N number?
    For equations of these particular form, yes. This does not work, for example, for \binom{N}{3}=\binom{N+1}{7}. Also, it helps to understand why the sum of the two bottom numbers is the solution. This was explained in posts #7 and #8.

    Quote Originally Posted by uperkurk View Post
    Ok so here I see how you have come to 15, just add the two numbers together, I'm not all that familiar with mathematic notation, what does the . mean between 8.7.6.5
    It's the multiplication sign.

    Also, the word is "formula," not "formular."
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    re: (n choose k) How can I work this out?

    In the exam I'll most likely have to provide some working out, I'll just say what you said

    k = N - j or N = k + j
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