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Math Help - Finding n given the ratio of nCr to another ratio

  1. #1
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    Talking Finding n given the ratio of nCr to another ratio

    In the expansion of (2+3x)^n the coefficients of x^3 and x^4 are in the raio 8:15. Find n.

    T_3 = \binom{n}{3}2^2(3x)^3
    T_4 = \binom{n}{4}2^3(3x)^4

    ratios are in 8:15, therefore,
     \frac{108\binom{n}{4}}{648\binom{n}{5}} = \frac{8}{15}
     \frac{\binom{n}{4}}{6\binom{n}{5}} \frac{8}{15}
     \frac{\frac{n!}{(n-4)!4!}}{\frac{6n!}{(n-5)!5!}} = \frac{8}{15}
    \frac{48n!}{(n-5)!5!} = \frac{15n!}{(n-4)!4!}
     \frac{2n!}{5(n-5)!} = \frac{5n!}{(n-4)!}
    \frac{2n!}{5(n-5)!} = \frac{5n!}{8(n-4)(n-5)!}
    \frac{2n!}{5(n-5)!} = \frac{5n!}{8(n-4)(n-5)!}
     2n!.8.(n-4) = 5n!.5
     16n - 64 = 25
     16n= 89
     n = 5.5625

    BUT OMG the answer is 8!!!!!!!!!!!!!!!!

    WHY!! what's wrong with my working out.
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  2. #2
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    Hello, differentiate!

    In the expansion of (2+3x)^n, the coefficients of x^3 and x^4 are in the raio 8:15.
    \text{Find }n.

    T_3 = \binom{n}{3}2^2(3x)^3,\quad T_4 = \binom{n}{4}2^3(3x)^4 . . . . . These are wrong
    The two terms are:

    . . T_3 \:=\: {n\choose3}2^{n-3}(3x)^3

    . . T_4 \:=\: {n\choose4}2^{n-4}(3x)^4


    Then: . \dfrac{{n\choose3}2^{n-3}(3^3)}{{n\choose4}2^{n-4}(3^4)} \;=\;\frac{8}{15} \quad\Rightarrow\quad \frac{\dfrac{n(n-1)(n-2)}{3!}}{\dfrac{n(n-1)(n-2)(n-3)}{4!}}\cdot\frac{2}{3} \:=\:\frac{8}{15}

    . . . . . . \frac{4}{n-3}\cdot\frac{2}{3} \:=\:\frac{8}{15} \quad\Rightarrow\quad \frac{1}{n-3} \:=\:\frac{1}{5}<br />


    . . . . . . 5 \:=\:n-3 \quad\Rightarrow\quad \boxed{n \:=\:8}

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