Question

If $\dfrac{1}{{{}^5{C_r}}}{\text{ }} + {\text{ }}\dfrac{1}{{{}^6{C_r}}}{\text{ }} = {\text{ }}\dfrac{1}{{{}^4{C_r}}},$ then the value of r equals to

Hint:- Value of r cannot be more than 4 because in the given equation the minimum value of n involved in ${}^n{C_r}$ is 4.

As we know that for any positive number $n$ and non-negative number $r$
$\Rightarrow {}^n{C_r}{\text{ }} = {\text{ }}\dfrac{{n!}}{{r!(n{\text{ }} - {\text{ }}r)!}}$ where ${\text{n }} \geqslant {\text{ r}}$.
And as we see that there is ${}^4{C_r}$ in the given expression, so, ${\text{r }} \leqslant {\text{ }}4$.
$\Rightarrow$Given equation is $\dfrac{1}{{{}^5{C_r}}}{\text{ }} + {\text{ }}\dfrac{1}{{{}^6{C_r}}}{\text{ }} = {\text{ }}\dfrac{1}{{{}^4{C_r}}},$ (1)
So, taking LCM of LHS of equation (1), we get
$\Rightarrow \dfrac{{{}^6{C_r}{\text{ }} + {\text{ }}{}^5{C_r}}}{{{}^6{C_r}.{}^5{C_r}}}{\text{ }} = {\text{ }}\dfrac{1}{{{}^4{C_r}}}$
Now, for solving above equation, letâ€™s expand ${}^4{C_r},{\text{ }}{}^5{C_r}$ and ${}^6{C_r}$.
So, above equation becomes,
$\Rightarrow \dfrac{{\dfrac{{6!}}{{r!(6{\text{ }} - {\text{ }}r)!}}{\text{ }} + {\text{ }}\dfrac{{5!}}{{r!(5{\text{ }} - {\text{ }}r)!}}}}{{\dfrac{{6!}}{{r!(6{\text{ }} - {\text{ }}r)!}}*\dfrac{{5!}}{{r!(5{\text{ }} - {\text{ }}r)!}}}}{\text{ = }}\dfrac{1}{{\dfrac{{4!}}{{r!(4{\text{ }} - {\text{ }}r)!}}}}$
On manipulating above equation, it can be written as,
$\Rightarrow \dfrac{{\dfrac{{6*5!}}{{r!*(6{\text{ }} - {\text{ }}r)*(5{\text{ }} - {\text{ }}r)!}}{\text{ }} + {\text{ }}\dfrac{{5!}}{{r!(5{\text{ }} - {\text{ }}r)!}}}}{{\dfrac{{6!}}{{r!(6{\text{ }} - {\text{ }}r)!}}*\dfrac{{5!}}{{r!(5{\text{ }} - {\text{ }}r)!}}}}{\text{ = }}\dfrac{1}{{\dfrac{{4!}}{{r!(4{\text{ }} - {\text{ }}r)!}}}}$
Taking $\dfrac{{5!}}{{r!(5{\text{ }} - {\text{ }}r)!}}$ common from the numerator and denominator, we get
$\Rightarrow \dfrac{{\left( {\dfrac{6}{{6{\text{ }} - {\text{ }}r}}{\text{ }} + {\text{ }}1} \right)}}{{\dfrac{{6!}}{{r!(6{\text{ }} - {\text{ }}r)!}}}}{\text{ }} = {\text{ }}\dfrac{{\left( {\dfrac{{12{\text{ }} - {\text{ }}r}}{{6{\text{ }} - {\text{ }}r}}} \right)}}{{\dfrac{{6!}}{{r!(6{\text{ }} - {\text{ }}r)!}}}}{\text{ }} = {\text{ }}\dfrac{1}{{\dfrac{{4!}}{{r!(4{\text{ }} - {\text{ }}r)!}}}}$
Solving above equation, we get
$\Rightarrow \left( {\dfrac{{12{\text{ }} - {\text{ }}r}}{{6{\text{ }} - {\text{ }}r}}} \right){\text{ }} = {\text{ }}\dfrac{{\dfrac{{6!}}{{r!.\left( {6{\text{ }} - {\text{ }}r} \right)!}}}}{{\dfrac{{4!}}{{r!.\left( {4{\text{ }} - {\text{ }}r} \right)!}}}}{\text{ }} = {\text{ }}\dfrac{{6*5}}{{\left( {6{\text{ }} - {\text{ }}r} \right)\left( {5{\text{ }} - {\text{ }}r} \right)}}$
So, from above equation, we get
$\Rightarrow \left( {\dfrac{{12{\text{ }} - {\text{ }}r}}{{6{\text{ }} - {\text{ }}r}}} \right){\text{ }} = {\text{ }}\dfrac{{30}}{{\left( {6{\text{ }} - {\text{ }}r} \right)\left( {5{\text{ }} - {\text{ }}r} \right)}}$ â€¦â€¦â€¦... (2)
Now, cross-multiplying equation (2),
$\Rightarrow \left( {12{\text{ }} - {\text{ }}r} \right)\left( {5{\text{ }} - {\text{ }}r} \right){\text{ }} = {\text{ }}30$ â€¦â€¦â€¦...(3)
On solving equation 3 it becomes,
$\Rightarrow {r^2}{\text{ }} - {\text{ }}17r{\text{ }} + {\text{ }}30{\text{ }} = {\text{ }}0$
Now, we had to solve the above equation to get different values of r possible.
$\Rightarrow \left( {r{\text{ }} - {\text{ }}15} \right)\left( {r{\text{ }} - {\text{ }}2} \right){\text{ }} = {\text{ }}0$
Hence r can be 15 or 2 but,
As we have said earlier that ${\text{r }} \leqslant {\text{ 4}}$,
$\Rightarrow$So, the value of r will be. ${\text{r }} = {\text{ }}2$.

Note:- Whenever we come up with this type of problem then efficient and easiest way to get the required value of r is by changing the given equation to a polynomial equation of r by using relation ${}^n{C_r}{\text{ }} = {\text{ }}\dfrac{{n!}}{{r!(n{\text{ }} - {\text{ }}r)!}}$. And then we can get the value of r by after solving the polynomial equation.