Question

If ${}^n{C_{r - 1}} = 10,{}^n{C_r} = 45$ and ${}^n{C_{r + 1}} = 120$ then $r$ equals
A. $1$
B. $2$
C. $3$
D. $4$

Answer 1

Hint: In this problem, we will use the properties of combination. We will use the property $\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}} = \dfrac{{n - r + 1}}{r}$ of combination. After using this property, we will get two equations in two variables. We will solve these two equations by a simple elimination method.


Complete step by step solution
In this problem, it is given that
${}^n{C_{r - 1}} = 10 \cdots \cdots \left( 1 \right)$
${}^n{C_r} = 45 \cdots \cdots \left( 2 \right)$
${}^n{C_{r + 1}} = 120 \cdots \cdots \left( 3 \right)$
Now we are going to use the property of combination which is given by $\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}} = \dfrac{{n - r + 1}}{r} \cdots \cdots \left( 4 \right)$.
Now we are going to substitute values from equation $\left( 1 \right)$ and $\left( 2 \right)$ in equation $\left( 4 \right)$. Therefore, we get
$
  \dfrac{{45}}{{10}} = \dfrac{{n - r + 1}}{r} \\
   \Rightarrow \dfrac{9}{2} = \dfrac{{n - r + 1}}{r} \\
   \Rightarrow 9r = 2\left( {n - r + 1} \right) \\
   \Rightarrow 9r = 2n - 2r + 2 \\
   \Rightarrow 9r + 2r - 2n = 2 \\
   \Rightarrow 11r - 2n = 2 \cdots \cdots \left( 5 \right) \\
 $
Let us replace $r$ by $r + 1$ in equation $\left( 4 \right)$. Therefore, we get
$
  \dfrac{{{}^n{C_{r + 1}}}}{{{}^n{C_{r + 1 - 1}}}} = \dfrac{{n - \left( {r + 1} \right) + 1}}{{r + 1}} \\
   \Rightarrow \dfrac{{{}^n{C_{r + 1}}}}{{{}^n{C_r}}} = \dfrac{{n - r - 1 + 1}}{{r + 1}} \\
   \Rightarrow \dfrac{{{}^n{C_{r + 1}}}}{{{}^n{C_r}}} = \dfrac{{n - r}}{{r + 1}} \cdots \cdots \left( 6 \right) \\
 $
Now we are going to substitute values from equation $\left( 2 \right)$ and $\left( 3 \right)$ in equation $\left( 6 \right)$. Therefore, we get
$
  \dfrac{{120}}{{45}} = \dfrac{{n - r}}{{r + 1}} \\
   \Rightarrow \dfrac{8}{3} = \dfrac{{n - r}}{{r + 1}} \\
   \Rightarrow 8\left( {r + 1} \right) = 3\left( {n - r} \right) \\
   \Rightarrow 8r + 8 = 3n - 3r \\
   \Rightarrow 8r + 3r - 3n = - 8 \\
   \Rightarrow 11r - 3n = - 8 \cdots \cdots \left( 7 \right) \\
 $
Now we have two equations in two variables $n$ and $r$. To find the value of $r$, we will eliminate $n$ from the equations $\left( 5 \right)$ and $\left( 7 \right)$. For this, we will multiply by number $3$ on both sides of equation $\left( 5 \right)$ and also multiply by number $2$ on both sides of equation $\left( 7 \right)$. Therefore, we get
$
  33r - 6n = 6 \cdots \cdots \left( 8 \right) \\
  22r - 6n = - 16 \cdots \cdots \left( 9 \right) \\
 $
Now we will subtract equation $\left( 9 \right)$ from equation $\left( 8 \right)$. Therefore, we get
$
  \left( {33r - 6n} \right) - \left( {22r - 6n} \right) = 6 - \left( { - 16} \right) \\
   \Rightarrow 33r - 6n - 22r + 6n = 6 + 16 \\
   \Rightarrow 11r = 22 \\
   \Rightarrow r = \dfrac{{22}}{{11}} \\
   \Rightarrow r = 2 \\
 $
Therefore, option B is correct.


Note:When solving these type of problems,make use of the appropriate property/formula of combination in accordance with the result which has to be found out/proved