Question

What is the ratio of $^{20}{C_r}$ to $^{25}{C_r}$ when each of them has the greatest value possible?
A) $\dfrac{{143}}{{4000}}$
B) $\dfrac{{143}}{{4025}}$
C) $\dfrac{4}{5}$
D) None of these

Answer 1

Hint: First we have to find for what value $^{20}{C_r}$ and $^{25}{C_r}$ gives the highest value. So, for a term like $^n{C_r}$, if $n$ is even, then the highest term is when $r = \dfrac{n}{2}$. And, if $n$ is odd, the highest term is when $r = \dfrac{{n + 1}}{2}{\text{ or }}\dfrac{{n - 1}}{2}$, both gives the same value. So, by finding the values, we can find the ratio $^{20}{C_r}$ to $^{25}{C_r}$, which is required by us.

Complete answer:
So, here given to us, the terms for combination are, $^{20}{C_r}$ and $^{25}{C_r}$.
Therefore, in $^{20}{C_r}$, $n = 20$, which is even.
So, we know, if $n$ is even, then the highest term is when $r = \dfrac{n}{2}$.
Therefore, $r = \dfrac{{20}}{2} = 10$.
So, $^{20}{C_r}{ = ^{20}}{C_{10}} = \dfrac{{20!}}{{10!\left( {20 - 10} \right)!}}$
${ \Rightarrow ^{20}}{C_{10}} = \dfrac{{20!}}{{10!.10!}}$
Also, in $^{25}{C_r}$, $n = 25$, which is odd.
So, we know, if $n$ is odd, then the highest term is when $r = \dfrac{{n + 1}}{2}{\text{ or }}\dfrac{{n - 1}}{2}$
Therefore, $r = \dfrac{{25 + 1}}{2} = \dfrac{{26}}{2} = 13$.
So, $^{25}{C_r}{ = ^{25}}{C_{13}} = \dfrac{{25!}}{{13!\left( {25 - 13} \right)!}}$
${ \Rightarrow ^{25}}{C_{13}} = \dfrac{{25!}}{{13!.12!}}$
[If we considered, $r = \dfrac{{n - 1}}{2}$, then also, the ultimate result would have been same]
Therefore, Maximum value of $\left[ {\dfrac{{^{20}{C_r}}}{{^{25}{C_r}}}} \right]$
$ = \dfrac{{^{20}{C_{10}}}}{{^{25}{C_{13}}}}$
$ = \dfrac{{\dfrac{{20!}}{{10!.10!}}}}{{\dfrac{{25!}}{{13!.12!}}}}$
Simplifying it, we get,
$ = \dfrac{{20!}}{{10!.10!}}.\dfrac{{13!.12!}}{{25!}}$
Now, we know, $n! = n.\left( {n - 1} \right).\left( {n - 2} \right).....2.1$.
So, using this property, we get,
$ = \dfrac{{20!}}{{10!.10!}}.\dfrac{{(13.12.11.10!).(12.11.10!)}}{{(25.24.23.22.21.20!)}}$
Therefore, simplifying this, we get,
$ = \dfrac{1}{{1.1}}.\dfrac{{13.12.11.12.11}}{{25.24.23.22.21}}$
Converting the given fraction into simplest form, we get,
$ = \dfrac{{13.11}}{{25.23.7}}$
$ = \dfrac{{143}}{{4025}}$
Therefore, the ratio of $^{20}{C_r}$ to $^{25}{C_r}$ is $\dfrac{{143}}{{4025}}$.

Thus, the correct option is B.

Note:
Permutations and combinations have many formulas that can be used to convert them from one form to another. This makes it easier to solve. Permutations and combinations have many real life applications in daily life, computer science, data science, bioengineering and many more.