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Thread: The Binomial Theorem/Factorials?

  1. #1
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    The Binomial Theorem/Factorials?

    Im a little confused on how to do this problem.

    Use the definition of [n] to show that [ 6 ] + [6] = [7] for all integers 1≤r≤6
    [r ] [r-1] [r] = [ r]

    The formula/definition is n!/r!(n - r!) am i correct? Help or an explanation of how to use this formula would be great! Thanks!!
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  2. #2
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    firstly it is important to know that

    $\displaystyle n! = n\times (n-1) \times (n-2) \times \cdots \times 3 \times 2 \times 1
    $

    for example $\displaystyle 4! =4\times 3 \times 2 \times 1 \times = 24 $

    and $\displaystyle 25! = 25 \times 24 \times \dots 3 \times 2 \times 1 \times$ = a massive number.

    For calculating $\displaystyle \binom{n}{r} =\frac{n!}{r!(n - r!)} $

    make sure you cancel as many terms as you can to make life easy

    i.e. $\displaystyle \binom{6}{3} =\frac{6!}{3!( 6- 3!)}=\frac{6!}{3!( 3!)} =\frac{6\times 5\times 4\times 3!}{3!( 3!)} = \frac{6\times 5\times 4}{3!} = \frac{120}{6}= 20$
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  3. #3
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    Quote Originally Posted by pickslides View Post
    firstly it is important to know that

    $\displaystyle n! = n\times (n-1) \times (n-2) \times \cdots \times 3 \times 2 \times 1
    $

    for example $\displaystyle 4! =4\times 3 \times 2 \times 1 \times = 24 $

    and $\displaystyle 25! = 25 \times 24 \times \dots 3 \times 2 \times 1 \times$ = a massive number.

    For calculating $\displaystyle \binom{n}{r} =\frac{n!}{r!(n - r!)} $

    make sure you cancel as many terms as you can to make life easy

    i.e. $\displaystyle \binom{6}{3} =\frac{6!}{3!( 6- 3!)}=\frac{6!}{3!( 3!)} =\frac{6\times 5\times 4\times 3!}{3!( 3!)} = \frac{6\times 5\times 4}{3!} = \frac{120}{6}= 20$

    [6] + [6] = [7] for all integers 1≤r≤6.
    [r-1] [r] [r]

    I think i understand what your saying, however i am still confused as how to show the proof in the above statement...
    my work:
    6!/(r!-1(6-r!-1)) + 6!/(r!(6-r!)) = 7!/r!(7-r!)

    Is this the correct way of setting this up? If so, where do i go from here?
    Im confused, but thanks a lot for your time!
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  4. #4
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    Quote Originally Posted by Sxon View Post
    [6] + [6] = [7] for all integers 1≤r≤6.
    [r-1] [r] [r]

    I think i understand what your saying, however i am still confused as how to show the proof in the above statement...
    my work:
    6!/(r!-1(6-r!-1)) + 6!/(r!(6-r!)) = 7!/r!(7-r!)

    Is this the correct way of setting this up? If so, where do i go from here?
    Im confused, but thanks a lot for your time!
    Why don't you just evaluate $\displaystyle \left(^6_r\right)$ for all $\displaystyle r \in [1, 6]$?


    Draw Pascal's Triangle and read off the 6th line... (the 1 on the top is the ZERO line...)
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  5. #5
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    Hello Sxon
    Quote Originally Posted by Sxon View Post
    Im a little confused on how to do this problem.

    Use the definition of [n] to show that [ 6 ] + [6] = [7] for all integers 1≤r≤6
    [r ] [r-1] [r] = [ r]

    The formula/definition is n!/r!(n - r!) am i correct? Help or an explanation of how to use this formula would be great! Thanks!!
    I think you mean to prove that $\displaystyle \binom6r + \binom{6}{r-1}= \binom7r$

    First, note that $\displaystyle r! = (r-1)!\times r$

    Then

    $\displaystyle \binom6r + \binom{6}{r-1} = \frac{6!}{r!(6-r)!}+\frac{6!}{(r-1)!(6-r+1)!}$

    $\displaystyle = \frac{6!}{(r-1)!\times r\times(6-r)!}+\frac{6!}{(r-1)!\times(6-r)!\times (6-r+1)}$

    $\displaystyle = \frac{6!\times (6-r+1)}{(r-1)!\times r\times(6-r)!\times(6-r+1)}+\frac{6!\times r}{(r-1)!\times(6-r)!\times (6-r+1)\times r}$

    $\displaystyle = \frac{6!\times (6-r+1)+6!\times r}{(r-1)!\times(6-r)!\times (6-r+1)\times r}$

    $\displaystyle = \frac{6!(6-r+1+r)}{r!(6-r+1)!}$

    $\displaystyle = \frac{6!\times 7}{r!(7-r)!}$

    $\displaystyle = \frac{7!}{r!(7-r)!}$

    $\displaystyle =\binom7r$

    Note that you can replace $\displaystyle 6$ throughout this proof by $\displaystyle n$, and get $\displaystyle \binom{n}{r}+\binom{n}{r-1}= \binom{n+1}{r}$

    Grandad
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