Question

We have combinations $^n{C_{r - 1}} = 36{,^n}{C_r} = 84$ and $^n{C_{r + 1}} = 126$ then $r$ is equal to
A). $1$
B). $2$
C). $3$
D). None of these

Answer 1

Here we are asked to find the value $r$ from the given data. Here we will use the combination formula to find the value of $r$. First, we will expand given terms using a combination formula and form equations, and solve them to find the value $r$.
Formula Used: Formula that we need to know:
$^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$

Complete step-by-step solution:
It is given that $^n{C_{r - 1}} = 36{,^n}{C_r} = 84$ and $^n{C_{r + 1}} = 126$. We aim to find the value of$r$.
We know the formula for $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$. Using this let us expand the given terms.
$^n{C_{r - 1}} = \dfrac{{n!}}{{r - 1!\left( {n - r + 1} \right)!}} = 36...........(1)$
$^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} = 84...............(2)$
$^n{C_{r + 1}} = \dfrac{{n!}}{{r + 1!\left( {n - r - 1} \right)!}} = 126...........(3)$
Now let us solve these equations to find the value of $r$.
Dividing equation $(2)$ by $(1)$ we get
$\dfrac{{n!}}{{r!\left( {n - r} \right)!}} \times \dfrac{{r - 1!\left( {n - r + 1} \right)!}}{{n!}} = \dfrac{{84}}{{36}}$
Let us simplify the above expression.
$\dfrac{{\left( {n - r + 1} \right)}}{r} = \dfrac{{84}}{{36}}$
On further simplification we get
$\dfrac{{\left( {n - r + 1} \right)}}{r} = \dfrac{{84}}{{36}}$
$\dfrac{{\left( {n - r + 1} \right)}}{r} = \dfrac{7}{3}$
On cross multiplying the above, we get
$3\left( {n - r + 1} \right) = 7r$
$\Rightarrow 3n - 3r + 3 = 7r$
$\Rightarrow 10r = 3n + 3..........(4)$
Now let us divide the equation$(3)$ by $(2)$ we get
$\dfrac{{n!}}{{r + 1!\left( {n - r - 1} \right)!}} \times \dfrac{{r!\left( {n - r} \right)!}}{{n!}} = \dfrac{{126}}{{84}}$
On simplifying the above expression, we get
$\dfrac{{n!}}{{\left( {r + 1} \right)r!\left( {n - r - 1} \right)!}} \times \dfrac{{r!\left( {n - r} \right)\left( {n - r - 1} \right)!}}{{n!}} = \dfrac{{126}}{{84}}$
On further simplification we get
$\dfrac{{n - r}}{{r + 1}} = \dfrac{3}{2}$
On cross multiplying the above, we get
$2\left( {n - r} \right) = 3\left( {r + 1} \right)$
$\Rightarrow 2n - 2r = 3r + 3$
$\Rightarrow 5r = 2n - 3.........(5)$
Now let us divide the equation $(4)$ by $(5)$we get
$\dfrac{{10r}}{{5r}} = \dfrac{{3n + 3}}{{2n - 3}}$
On simplifying the above expression, we get
$2 = \dfrac{{3n + 3}}{{2n - 3}}$
$\Rightarrow 2\left( {2n - 3} \right) = 3n + 3$
$\Rightarrow 4n - 6 = 3n + 3$
$\Rightarrow n = 9$
Now let us substitute the value $n = 9$ the equation $(4)$ we get
$(4) \Rightarrow 10r = 3(9) + 3$
On simplifying the above expression, we get
$\Rightarrow 10r = 30$
$\Rightarrow r = 3$
Thus, we got the value$r = 3$. Now let us see the options, option (a) $1$ is an incorrect option as we got $r = 3$ in our calculation.
Option (b) $2$ is an incorrect option as we got $r = 3$ in our calculation.
Option (c) $3$ is the correct option as we got the same answer in the calculation above.
Option (d) None of these is an incorrect answer as we got option (c) as a correct option.
Hence, option (c) $3$ is the correct answer.

Note: A factorial of any number $n$ is written as $n! = n(n - 1)(n - 2)(n - 3)...3.2.1$. Thus, the factorial expansion can also be written as $n! = n(n - 1)!$ since $(n - 1)! = (n - 1)(n - 2)(n - 3)...3.2.1$. We have used this form of expansion in our calculation above.