Question

If the ratio ${}^{28}{C_{2r}}:{}^{24}{C_{2r - 4}} = 225:11$ , then find $r$ .

Answer 1

Hint:Equate both the sides after expanding the combinations using their definition. Cancel out the common terms from numerator and denominator. Now compare the obtained fraction with the given right side of the equation, i.e. $\dfrac{{225}}{{11}}$ . Both numerator and denominator should be compared to get the required value.

Complete step-by-step answer:
Let’s first understand the concept of combinations. A combination is a mathematical technique that determines the number of possible arrangements in a collection of items where the order of the selection does not matter. In combinations, you can select the items in any order.
Now suppose we want to choose $k$ objects from $n$ objects, then the number of combinations of $k$ objects chosen from $n$ objects is denoted by ${}^n{C_k}$ or $\left( {\begin{array}{*{20}{c}}
  n \\
  k
\end{array}} \right)$ , it follows that:
$ \Rightarrow \left( {\begin{array}{*{20}{c}}
  n \\
  k
\end{array}} \right) = {}^n{C_k} = \dfrac{{n!}}{{k!\left( {n - k} \right)!}}$
Let’s take the left side of the given equation and try to simplify it using the definition of combinations:
$ \Rightarrow \dfrac{{{}^{28}{C_{2r}}}}{{{}^{24}{C_{2r - 4}}}} = \dfrac{{\dfrac{{28!}}{{2r!\left( {28 - 2r} \right)!}}}}{{\dfrac{{24!}}{{\left( {2r - 4} \right)!\left( {24 - 2r + 4} \right)!}}}}$
Now, we can arrange these factorials in both numerator and denominator:
$ \Rightarrow \dfrac{{{}^{28}{C_{2r}}}}{{{}^{24}{C_{2r - 4}}}} = \dfrac{{\dfrac{{28!}}{{2r!\left( {28 - 2r} \right)!}}}}{{\dfrac{{24!}}{{\left( {2r - 4} \right)!\left( {24 - 2r + 4} \right)!}}}} = \dfrac{{28!\left( {2r - 4} \right)!\left( {28 - 2r} \right)!}}{{2r!\left( {28 - 2r} \right)! \times 24!}}$
After this, we can cancel out the common factorial of $\left( {28 - 2r} \right)$ from numerator and denominator:
$ \Rightarrow \dfrac{{{}^{28}{C_{2r}}}}{{{}^{24}{C_{2r - 4}}}} = \dfrac{{28!\left( {2r - 4} \right)!\left( {28 - 2r} \right)!}}{{2r!\left( {28 - 2r} \right)! \times 24!}} = \dfrac{{28!\left( {2r - 4} \right)!}}{{2r! \times 24!}}$
We can now expand $28!$ and $2r!$ using the definition $n! = n \times \left( {n - 1} \right)! = n\left( {n - 1} \right) \times \left( {n - 2} \right)!$ as:
$ \Rightarrow \dfrac{{{}^{28}{C_{2r}}}}{{{}^{24}{C_{2r - 4}}}} = \dfrac{{28!\left( {2r - 4} \right)!}}{{2r! \times 24!}} = \dfrac{{28 \times 27 \times 26 \times 25 \times 24!\left( {2r - 4} \right)!}}{{2r \times \left( {2r - 1} \right) \times \left( {2r - 2} \right) \times \left( {2r - 3} \right) \times \left( {2r - 4} \right)!24!}}$
Now, we can simplify it further as:
\[ \Rightarrow \dfrac{{{}^{28}{C_{2r}}}}{{{}^{24}{C_{2r - 4}}}} = \dfrac{{28 \times 27 \times 26 \times 25 \times 24!\left( {2r - 4} \right)!}}{{2r \times \left( {2r - 1} \right) \times \left( {2r - 2} \right) \times \left( {2r - 3} \right) \times \left( {2r - 4} \right)!24!}} = \dfrac{{28 \times 27 \times 26 \times 25}}{{2r \times \left( {2r - 1} \right) \times \left( {2r - 2} \right) \times \left( {2r - 3} \right)}}\]
Let’s substitute this into the given equation, we get:
\[ \Rightarrow \dfrac{{{}^{28}{C_{2r}}}}{{{}^{24}{C_{2r - 4}}}} = \dfrac{{28 \times 27 \times 26 \times 25}}{{2r \times \left( {2r - 1} \right) \times \left( {2r - 2} \right) \times \left( {2r - 3} \right)}} = \dfrac{{225}}{{11}}\]
Therefore, from the above equation, we can conclude that the denominator of the left side must have $11$ as one of its factors. There is no multiple of $11$ in numerator. Since $11$ is a prime number and cannot be cancelled out with a factor in the numerator.
Thus, one of the numbers from \[2r,\left( {2r - 1} \right),\left( {2r - 2} \right){\text{ and }}\left( {2r - 3} \right)\] in denominator must be $11$.
Also here \[2r{\text{ and }}\left( {2r - 2} \right){\text{ }}\] are multiples of two and hence an even number. But $11$ is a prime, so it should be one of $\left( {2r - 1} \right){\text{ or }}\left( {2r - 3} \right)$ .
If $2r - 1 = 11$ then $ \Rightarrow r = \dfrac{{11 + 1}}{2} = 6$ and if $2r - 3 = 11$ then $ \Rightarrow r = \dfrac{{11 + 3}}{2} = 7$
Thus, we can say the value of $r$ is either $6$ or $7$.
For $r = 6$ , we get:
\[ \Rightarrow \dfrac{{{}^{28}{C_{2r}}}}{{{}^{24}{C_{2r - 4}}}} = \dfrac{{28 \times 27 \times 26 \times 25}}{{2r \times \left( {2r - 1} \right) \times \left( {2r - 2} \right) \times \left( {2r - 3} \right)}} = \dfrac{{28 \times 27 \times 26 \times 25}}{{12 \times 11 \times 10 \times 9}} = \dfrac{{455}}{{11}} \ne \dfrac{{225}}{{11}}\]
So, for $r = 7$ , we get:
\[ \Rightarrow \dfrac{{{}^{28}{C_{2r}}}}{{{}^{24}{C_{2r - 4}}}} = \dfrac{{28 \times 27 \times 26 \times 25}}{{2r \times \left( {2r - 1} \right) \times \left( {2r - 2} \right) \times \left( {2r - 3} \right)}} = \dfrac{{28 \times 27 \times 26 \times 25}}{{14 \times 13 \times 12 \times 11}} = \dfrac{{225}}{{11}}\]
Hence, we get the left side of the equation equal to the right side of the equation.
Therefore value of $r$ is $7$

Note:Notice that the use of the definition of the combination ${}^n{C_r}$ was the most crucial part of the solution. An alternative approach to this problem can be equating both sides as:\[ \Rightarrow 2r\left( {2r - 1} \right)\left( {2r - 2} \right)\left( {2r - 3} \right) = \dfrac{{28 \times 27 \times 26 \times 25 \times 11}}{{225}} = 14 \times 13 \times 12 \times 11\] .Now putting $2r = 14$ for finding the values of $r$ .