Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

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

seo-qna
Last updated date: 22nd Mar 2024
Total views: 415.8k
Views today: 5.15k
MVSAT 2024
Answer
VerifiedVerified
415.8k+ views
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.