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Question: If the value of \(^n{C_6}{:^{n - 3}}{C_3} = 33:4,\) find the value of n....

If the value of nC6:n−3C3=33:4,^n{C_6}{:^{n - 3}}{C_3} = 33:4, find the value of n.

Explanation

Solution

Hint: Let's make use of the formula nCr=n!(n−r)!r!{n_{{C_r} = \dfrac{{n!}}{{(n - r)!r!}}}} $$$$ and solve this.

Complete step-by-step answer:
By making use of the formula nCr=n!(n−r)!r!{n_{{C_r} = \dfrac{{n!}}{{(n - r)!r!}}}}
We can write nC6=n!(n−6)!6!{n_{{C_6} = \dfrac{{n!}}{{(n - 6)!6!}}}}
n−3C3=n!(n−6)!3!n - {3_{{C_3} = \dfrac{{n!}}{{(n - 6)!3!}}}}
So ,now we can write the ratio nC6n−3C3=334\dfrac{{^n{C_6}}}{{^{n - 3}{C_3}}} = \dfrac{{33}}{4}
Lets substitute the values of nC6^n{C_6} and nC3^n{C_3}
So, we get n!(n−6)!6!(n−3)!(n−6)!3!=334\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)!n! = n(n - 1)! and write

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

So, from this we can cancel out (n−3)!,(n−6)!(n - 3)!,(n - 6)! in the numerator and denominator
n(n−1)(n−2)6!×3!=334\dfrac{{n(n - 1)(n - 2)}}{{6!}} \times 3! = \dfrac{{33}}{4}

On shifting 6!6! and 3!3! to the right hand side, we get
n(n−1)(n−2)=11×3×5×2×3=11×10×9n(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)!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.