Question

If \[{}^{2n}{c_3}:{}^n{c_2} = 44:3\] then for which of the following value of \[r\] , the value of \[{}^n{c_r}\] will be \[15\]?
A. \[r = 3\]
B. \[r = 4\]
C. \[r = 6\]
D. \[r = 5\]

Answer 1

Hint:In the given problem, we have to find the value of \[r\]. We will proceed with \[{}^{2n}{c_3}:{}^n{c_2} = 44:3\] and simplify to find the value of \[n\]. We will use the formula \[{}^n{c_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] in above and solve it to get the value of \[n\]. We will then compare \[{}^n{c_r}\] to \[{}^6{c_2} = 15\] to get the value of \[r\].

Complete step by step answer:
Consider the given question,
We are given , \[{}^{2n}{c_3}:{}^n{c_2} = 44:3\].
i.e. \[\dfrac{{{}^{2n}{c_3}}}{{{}^n{c_2}}} = \dfrac{{44}}{3}\]
Cross multiplying and then putting the value of \[{}^{2n}{c_3}\] and \[{}^n{c_2}\] we have,
\[ \Rightarrow 3 \times \dfrac{{2n!}}{{3!\left( {2n - 3} \right)!}} = 44 \times \dfrac{{n!}}{{2!\left( {n - 2} \right)!}}\]
On simplifying, we have
\[ \Rightarrow 3 \times \dfrac{{2n(2n - 1)(2n - 2)(2n - 3)!}}{{3!\left( {2n - 3} \right)!}} = 44 \times \dfrac{{n(n - 1)(n - 2)!}}{{2!\left( {n - 2} \right)!}}\]
Cancelling \[(2n - 3)!\] in LHS and \[(n - 2)!\] in RHS we have,
\[ \Rightarrow 3 \times \dfrac{{2n(2n - 1)(2n - 2)}}{{3!}} = 44 \times \dfrac{{n(n - 1)}}{{2!}}\]
Taking common in LHS and on simplifying we have,
\[ \Rightarrow 3 \times \dfrac{{2 \times 2n(2n - 1)(n - 1)}}{{3 \times 2 \times 1}} = 44 \times \dfrac{{n(n - 1)}}{{2 \times 1}}\]

On simplifying, by cancelling \[n(n - 1)\] both side and the equal terms we get
\[ \Rightarrow 2(2n - 1) = 22\]
On simplifying,
\[ \Rightarrow 4n - 2 = 22\]
Adding \[2\] both side, we get
\[ \Rightarrow 4n = 24\]
Dividing by \[4\] both side we get
\[ \Rightarrow n = 6\].
Now we have to find the value of \[r\], when \[{}^n{c_r} = 15\]
Consider, \[{}^n{c_r} = 15\]
We know that \[{}^6{c_2} = 15\]
Hence, \[{}^6{c_r} = {}^6{c_2}\]
Comparing, we have \[r = 2\] or \[4\].

Hence option B is correct.

Note:The factorial of any number is written as \[n! = n(n - 1)(n - 2)...........3 \times 2 \times 1\]. We can also write \[n! = n \times (n - 1)!\], this will help to cancel the terms easily. The process of selecting r things out of $n$ thing is given by \[{}^n{c_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]. The process of arranging r things out of n things is given by \[{}^n{p_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\]