Question

If \[ {}^nC_{r-1}=36,\qquad {}^nC_r=84,\qquad {}^nC_{r+1}=126, \] then the value of \(nr^2\) is

• A. \(243\)
• B. \(9\)
• C. \(27\)
• D. \(81\)

Hint:

For consecutive binomial coefficients, use \[ \frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n-r+1}{r} \] and \[ \frac{{}^nC_{r+1}}{{}^nC_r} = \frac{n-r}{r+1}. \] These formulas quickly determine \(n\) and \(r\).

Answer 1

The Correct Option is D

Step 1: Use the ratio of consecutive binomial coefficients.
We know that \[ \frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n-r+1}{r}. \] Substituting the given values, \[ \frac{84}{36} = \frac{n-r+1}{r}. \] \[ \frac{7}{3} = \frac{n-r+1}{r}. \] Thus, \[ 3n-3r+3=7r. \] \[ 3n+3=10r. \] \[ r=\frac{3n+3}{10}. \]

Step 2: Use the second ratio.
Also, \[ \frac{{}^nC_{r+1}}{{}^nC_r} = \frac{n-r}{r+1}. \] Substituting values, \[ \frac{126}{84} = \frac{n-r}{r+1}. \] \[ \frac{3}{2} = \frac{n-r}{r+1}. \] Hence, \[ 2(n-r)=3(r+1). \] \[ 2n-2r=3r+3. \] \[ 2n=5r+3. \]

Step 3: Solve for \(n\) and \(r\).
Using \[ r=\frac{3n+3}{10}, \] in \[ 2n=5r+3, \] we get \[ 2n=5\left(\frac{3n+3}{10}\right)+3. \] \[ 4n=3n+3+6. \] \[ n=9. \] Therefore, \[ r=\frac{3(9)+3}{10} =\frac{30}{10} =3. \]

Step 4: Find \(nr^2\).
\[ nr^2=9(3)^2 \] \[ =9\times9 \] \[ =81. \]

Step 5: Final conclusion.
Hence, \[ \boxed{81} \]