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# If the value of $^n{C_6}{:^{n - 3}}{C_3} = 33:4,$ find the value of n.  Answer Verified
Hint: Let's make use of the formula ${n_{{C_r} = \dfrac{{n!}}{{(n - r)!r!}}}}$  and solve this.

Complete step-by-step answer:
By making use of the formula ${n_{{C_r} = \dfrac{{n!}}{{(n - r)!r!}}}}$
We can write ${n_{{C_6} = \dfrac{{n!}}{{(n - 6)!6!}}}}$
$n - {3_{{C_3} = \dfrac{{n!}}{{(n - 6)!3!}}}}$
So ,now we can write the ratio $\dfrac{{^n{C_6}}}{{^{n - 3}{C_3}}} = \dfrac{{33}}{4}$
Lets substitute the values of $^n{C_6}$ and $^n{C_3}$
So, we get $\dfrac{{_{\dfrac{{n!}}{{(n - 6)!6!}}}}}{{_{\dfrac{{(n - 3)!}}{{(n - 6)!3!}}}}} = \dfrac{{33}}{4}$
Here, we will make use of the formula $n! = n(n - 1)!$ and write

$n! = n(n - 1)(n - 2)(n - 3)!$ in the numerator .

So, from this we can cancel out $(n - 3)!,(n - 6)!$ in the numerator and denominator
$\dfrac{{n(n - 1)(n - 2)}}{{6!}} \times 3! = \dfrac{{33}}{4}$

On shifting $6!$ and $3!$ to the right hand side, we get
$n(n - 1)(n - 2) = 11 \times 3 \times 5 \times 2 \times 3 = 11 \times 10 \times 9$
Therefore n=11.

Note: While expressing $n! = n(n - 1)!$ express it upto the term which is present in the denominator so that the terms in the numerator and denominator will get cancelled out and it would be easy to solve.
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