Question

If \[{}^{2n}{C_2}:{}^n{C_2} = 9:2\], and \[{}^n{C_r} = 10\] then find the value of \[r\].
A. \[1\]
B. \[2\]
C. \[4\]
D. \[5\]

Answer 1

Hint: In the given question, we need to find the solution set of \[r\]. For this, we will simplify the expression \[{}^{2n}{C_2}:{}^n{C_2} = 9:2\] to get the value of \[n\] . Also, using this and \[{}^n{C_r} = 10\], we will find the value of \[r\]to get the desired result.



Formula Used:The following formula used for solving the given question.
The formula of combination is given by, \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
Here, \[n\] is total number of things and \[r\] is number of things need to be selected from total things.



Complete step by step solution:We know that \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
Here, \[n\] is total number of things and \[r\] is number of things need to be selected from total things.
Now, we will have the expression \[{}^{2n}{C_2}:{}^n{C_2} = 9:2\].
 \[\dfrac{{{}^{2n}{C_2}}}{{{}^n{C_2}}} = \dfrac{{\dfrac{{2n!}}{{2!\left( {2n - 2} \right)!}}}}{{\dfrac{{n!}}{{2!\left( {n - 2} \right)!}}}}\]
By simplifying, we get
\[\dfrac{{{}^{2n}{C_2}}}{{{}^n{C_2}}} = \dfrac{{2n!}}{{2!\left( {2n - 2} \right)!}} \times \dfrac{{2!\left( {n - 2} \right)!}}{{n!}}\]
\[\dfrac{{{}^{2n}{C_2}}}{{{}^n{C_2}}} = \dfrac{{2\left( {n - 2} \right)!}}{{\left( {2n - 2} \right)!}}\]
This gives
\[\dfrac{9}{2} = \dfrac{{2\left( {2n - 1} \right)}}{{\left( {n - 1} \right)}}\]
Thus, we get
 \[9\left( {n - 1} \right) = 4\left( {2n - 1} \right)\]
\[9n - 9 = 8n - 4\]
By simplifying, we get
\[n = 5\]
Also, we have
\[{}^n{C_r} = 10\]
\[{}^5{C_r} = 10\]
By simplifying, we get
\[{}^5{C_r} = \dfrac{{5!}}{{r!\left( {5 - r} \right)!}} = 10\]
Finally, we have
\[5 - r = 3\]
By simplifying, we get
\[r = 2\]
Hence, if \[{}^{2n}{C_2}:{}^n{C_2} = 9:2\], and \[{}^n{C_r} = 10\] then the value of \[r\]is \[2\].



Option ‘B’ is correct

Additional Information:Combinations are ways to choose elements from a group in mathematics in which the order of the selection is irrelevant. Permutations and combinations can be jumbled up. The sequence in which the chosen components are chosen is crucial in permutations, too. Combinatorics is the study of combinations; however, mathematics as well as finance are two other fields that use combinations.



Note: Many students make mistake in calculation part as well as writing the combination rule. This is the only way, through which we can solve the example in simplest way. Also, it is necessary to find the appropriate value of \[n\]as the value of \[r\] is totally depends on it.