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: Here, first need to find value of n using \[{}^{2n}{C_3}:{}^n{C_2} = 44:3\] and then substitute value of n in \[{}^n{C_r} = 15\] and find value of r.

Formula used: \[{}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}\]
Here, \[n\] is the total number of items and there are \[r\] combinations.

Complete step by step solution: The data given is \[{}^{2n}{C_3}:{}^n{C_2} = 44:3\].
Apply a combination formula and simplify the expression.
\[\dfrac{{\left( {2n} \right)!}}{{\left( {2n - 3} \right)!3!}} \div \dfrac{{\left( n \right)!}}{{\left( {n - 2} \right)!2!}} = \dfrac{{44}}{3}\]
Further simplify the equation.
\[\dfrac{{\left( {2n} \right)!}}{{\left( {2n - 3} \right)!3!}} \times \dfrac{{\left( {n - 2} \right)!2!}}{{n!}} = \dfrac{{44}}{3}\]
Now expand the factorial.
\[\dfrac{{\left( {2n} \right)\left( {2n - 1} \right)\left( {2n - 2} \right)\left( {2n - 3} \right)!}}{{\left( {2n - 3} \right)!3!}} \times \dfrac{{\left( {n - 2} \right)!2!}}{{n\left( {n - 1} \right)\left( {n - 2} \right)!}} = \dfrac{{44}}{3}\]
Simplify the expression.
\[\dfrac{{\left( {2n} \right)\left( {2n - 1} \right)\left( {2n - 2} \right)}}{{3!}} \times \dfrac{{2!}}{{n\left( {n - 1} \right)}} = \dfrac{{44}}{3}\]
Now further simplify as follows.
\[\dfrac{{\left( {2n} \right)\left( {2n - 1} \right)\left( {2n - 2} \right)}}{{3n\left( {n - 1} \right)}} = \dfrac{{44}}{3}\]
Multiply by \[3\] on both sides of the equation.
\[\dfrac{{\left( {2n} \right)\left( {2n - 1} \right)\left( {2n - 2} \right)}}{{n\left( {n - 1} \right)}} = 44\]
Multiply by \[n\left( {n - 1} \right)\] on both sides of the equation.
\[\left( {2n} \right)\left( {2n - 1} \right)\left( {2n - 2} \right) = 44n\left( {n - 1} \right)\]
Now simplify the equation to find the value of \[n\]. Here isolate \[n\].
\[\begin{array}{l}\left( {2n - 1} \right)\left( {2n - 2} \right) = 22\left( {n - 1} \right)\\2\left( {n - 1} \right)\left( {2n - 1} \right) = 22\left( {n - 1} \right)\end{array}\]
Divide by \[\left( {n - 1} \right)\] on both sides of the equation.
\[2\left( {2n - 1} \right) = 22\]
Divide by \[2\] on both sides of the equation.
\[\left( {2n - 1} \right) = 11\]
Now isolate \[n\]
\[\begin{array}{l}2n - 1 = 11\\2n = 12\\n = 6\end{array}\]
To find correct value of \[r\], substitute \[n = 6\] in the \[{}^n{C_r} = 15\]
\[{}^6{C_r} = 15\]
Here we know that \[{}^6{C_2} = 15\] and \[{}^6{C_4} = 15\]. So, the value of \[r\] can be \[2\] or \[4\].
Hence from the given option we can say the value of \[r\] is \[4\].

Thus, Option (B) is correct.

Note: Common students consider that \[{}^{2n}{C_3} = 44\] and \[{}^n{C_2} = 3\]. This gives the wrong value of n that leads to the wrong value of r. This is not the correct way to find the value of r.