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Math Help - nCr 'show that' qu.

  1. #1
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    nCr 'show that' qu.

    Questions:

    a) Suppose that r and n are integers with 1 \le r \le n.

    Show that ^n\mathrm{C}_r + ^n\mathrm{C}_{r-1} = ^{n+1}\mathrm{C}_r

    b) If n is a positive integer,

    Show that 2^n= ^n\mathrm{C}_0 + ^n\mathrm{C}_1 + ^n\mathrm{C}_2 +......+ ^n\mathrm{C}_{n-1} + ^n\mathrm{C}_n

    __________________________________________________ _____________

    My attempt so far:

    a)  ^n\mathrm{C}_r + ^n\mathrm{C}_{r-1} = \frac{n!}{(n-r)!r!} + \frac{n!}{(n-(r-1))!(r-1)!}

     ^n\mathrm{C}_r + ^n\mathrm{C}_{r-1} = \frac{n! (n-r+1)!(r-1)! + n!(n-r)!r!}{(n-r)!r!(n-r+1)!(r-1)!}

    Ugh! Any ideas where I could go from here?

    b)  2^n = \sum (^n\mathrm{C}_n) = \sum \frac{n!}{(n-n)!n!} = \sum \frac{n!}{n!} = \sum 1

    which is clearly wrong.


    Please help! Thank you!
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  2. #2
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    \begin{array}{rcl} \frac{{n!}}{{\left( {n - r} \right)!r!}} + \frac{{n!}}{{\left( {n + 1 - r} \right)!\left( {r - 1} \right)!}} & = & \frac{{n!\left( {n + 1 - r} \right) + n!r}}{{\left( {n + 1 - r} \right)!r!}} \\ <br />
  & = & \frac{{n!\left( {n + 1} \right) - n!r}}{{\left( {n + 1 - r} \right)!r!}} \\ <br />
  & = & _{n + 1} C_r  \\  \end{array}<br />

    For the next one.
    \left( {x + y} \right)^n  = \sum\limits_{}^{} {_n C_k x^k y^{n - k} }
    Let x=1 & y=1.
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  3. #3
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    Quote Originally Posted by Plato View Post
    \begin{array}{rcl} \frac{{n!}}{{\left( {n - r} \right)!r!}} + \frac{{n!}}{{\left( {n + 1 - r} \right)!\left( {r - 1} \right)!}} & = & \frac{{n!\left( {n + 1 - r} \right) + n!r}}{{\left( {n + 1 - r} \right)!r!}} \\ <br />
& = & \frac{{n!\left( {n + 1} \right) - n!r}}{{\left( {n + 1 - r} \right)!r!}} \\ <br />
& = & _{n + 1} C_r \\ \end{array}<br />

    For the next one.
    \left( {x + y} \right)^n = \sum\limits_{}^{} {_n C_k x^k y^{n - k} }
    Let x=1 & y=1.
    For the first part, that's brilliant! I'll make a note of that trick. It's a lot easier than the way I tried to do it.


    And for (b), subtituting x=1, y=1 gives

    2^n = \sum nC_k

    How does the next step follow?

    Thanks for the help, once again!
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  4. #4
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    Quote Originally Posted by WWTL@WHL View Post
    How does the next step follow?
    Oh! Come on! EXPAND.
    \sum\limits_{k = 0}^N {_N C_k }  = _N C_0  + _N C_1  +  \cdots  + _N C_N
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  5. #5
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    Quote Originally Posted by Plato View Post
    Oh! Come on! EXPAND.
    \sum\limits_{k = 0}^N {_N C_k } = _N C_0 + _N C_1 + \cdots + _N C_N
    What?! In my textbook it says the sum of nCr goes to  _n C_1 + _n C_2+.....+ _N C_ {r-1} + _N C_r

    I'll have to check somewhere else. I used to be able to do this stuff without a second thought 2 years ago. It's been so long since I've touched stats.

    But thanks, Plato. You're a life-saver!
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  6. #6
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    Quote Originally Posted by WWTL@WHL View Post
    Show that 2^n= ^n\mathrm{C}_0 + ^n\mathrm{C}_1 + ^n\mathrm{C}_2 +......+ ^n\mathrm{C}_{n-1} + ^n\mathrm{C}_n
    That is exactly what I just posted.
    The text agrees.
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  7. #7
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    Quote Originally Posted by Plato View Post
    That is exactly what I just posted.
    The text agrees.
    No, I get that. It's just the \sum n C k goes to nC1 + nC2 +....+ nCn bit I didn't understand - if you understand what I'm saying. :

    All is well now, I had to look it up in one of my old high-school textbooks, and I get it now.

    Sorry for the confusion on the last bit. You were extremely helpful, yet again. Thanks.
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