Question

If \[4.{}^n{c_6} = 33.{}^{n - 3}{c_3}\] then \[n\] is equal to
A) 9
B) 10
C) 11
D) None of these

Answer 1

Hint: To find the required value of (n) we are applying the formula \[{}^n{c_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] , Where r is the number of objects to be selected from n number of distinct objects. After that we will calculate the required value for \[n\].

Complete step by step answer:
Formula used: \[{}^n{c_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] , Where r is the number of objects to be selected from n number of distinct objects.
According to the question
\[4.{}^n{c_6} = 33.{}^{n - 3}{c_3}\]
\[ \Rightarrow \dfrac{{{}^n{c_6}}}{{{}^{n - 3}{c_3}}} = \dfrac{{33}}{4}\]……………. (i)
Using the formula\[{}^n{c_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] in eqn (i), we get
\[ \Rightarrow \dfrac{{n!}}{{6!(n - 6)!}} \times \dfrac{{3!(n - 6)!}}{{n - 3!}} = \dfrac{{33}}{4}\]
\[ \Rightarrow \dfrac{{n!}}{{720(n - 6)!}} \times \dfrac{{6(n - 6)!}}{{n - 3!}} = \dfrac{{33}}{4}\]
\[ \Rightarrow \dfrac{{n!}}{{720}} \times \dfrac{6}{{n - 3!}} = \dfrac{{33}}{4}\]
\[ \Rightarrow \dfrac{{n!}}{{120}} \times \dfrac{1}{{n - 3!}} = \dfrac{{33}}{4}\]
\[ \Rightarrow n!{\text{ }} \times \dfrac{1}{{n - 3!}} = \dfrac{{33}}{4} \times 120\]
\[ \Rightarrow \dfrac{{n!}}{{(n - 3)!}} = 990\]…………………… (ii)
Using \[n! = n(n - 1)(n - 2)(n - 3)!\] in eqn (ii)
\[ \Rightarrow \dfrac{{n(n - 1)(n - 2)(n - 3)!}}{{(n - 3)!}} = 990\]
\[ \Rightarrow n(n - 1)(n - 2) = 990\]
\[ \Rightarrow n(n - 1)(n - 2) = 11 \times 10 \times 9\]……………… (iii)
On comparing both LHS and RHS of eqn (iii), we get

The value of \[n = 11\]

Therefore, Option (C) is the correct answer.

Note:
 To solve the permutation combination related question, we will have to understand the concept of arrangement and selection process and understand the meaning of formula related to this concept.
Some important properties:
\[{}^n{c_0} = 1\]
\[{}^n{c_1} = n\]
\[{}^n{c_n} = 1\]
One should always keep in mind that the value for \[n\] is always a positive integer.