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: Firstly, using the formula $\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}} = \dfrac{{n - r + 1}}{r}$ and substituting the appropriate values find a linear equation in terms of n and r.
Then, by replacing r by \[r + 1\] in the equation $\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}} = \dfrac{{n - r + 1}}{r}$ and substituting the appropriate values find another linear equation in terms of n and r.
Finally, solve these linear equations to find the value of r.

Complete step-by-step answer:
Here, it is given that, ${}^n{C_{r - 1}} = 10,{}^n{C_r} = 45$ and ${}^n{C_{r + 1}} = 120$ .
Now, to solve we will use the formula $\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}} = \dfrac{{n - r + 1}}{r}$ . … (1)
So, substituting the values ${}^n{C_{r - 1}} = 10,{}^n{C_r} = 45$ in the above equation.
 $
  \Rightarrow \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 2n - 2r - 9r + 2 = 0 \\
 $
 $\Rightarrow 2n - 11r + 2 = 0$ … (2)
Now, in the equation (1), substituting r = \[r + 1\] , 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}} \\
 $
Now, substituting the values of ${}^n{C_r} = 45$ and ${}^n{C_{r + 1}} = 120$ in the above equation 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 3n - 3r - 8r - 8 = 0 \\
 $
$\Rightarrow 3n - 11r - 8 = 0$ … (3)
Now, we will solve equations (2) and (3) to get the value of r.
On multiplying equation (2) by 3 and equation (3) by -2, we get
 $
  6n - 33r + 6 = 0 \\
   - 6n + 22r + 16 = 0 \\
 $
On solving the above pair of linear equations, we will get the required answer.
 $
  \Rightarrow - 11r + 22 = 0 \\
  \Rightarrow - 11\left( {r - 2} \right) = 0 \\
  \Rightarrow r - 2 = 0 \\
  \Rightarrow r = 2 \\
 $
Thus, we get the values of r as 2.
So, option (B) is correct.

Note: Proof that $\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}} = \dfrac{{n - r + 1}}{r}$ .
We know that, ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ and ${}^n{C_{r - 1}} = \dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r + 1} \right)!}}$
Now, taking ratio $\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}}$ , we get
 $
  \dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}} = \dfrac{{\dfrac{{n!}}{{r!\left( {n - r} \right)!}}}}{{\dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r + 1} \right)!}}}} \\
   = \dfrac{{\left( {r - 1} \right)!\left( {n - r + 1} \right)!}}{{r!\left( {n - r} \right)!}} \\
   = \dfrac{{\left( {r - 1} \right)!\left( {n - r + 1} \right)\left( {n - r} \right)!}}{{r\left( {r - 1} \right)!\left( {n - r} \right)!}} \\
   = \dfrac{{n - r + 1}}{r} \\
 $