Question:
If
\[
{}^{2n}C_3 : {}^{n}C_3 = 12:1,
\]
then \(n=\)
If
\[
{}^{2n}C_3 : {}^{n}C_3 = 12:1,
\]
then \(n=\)
Show Hint
When ratios of combinations are given, first write combinations in factorial form and simplify by cancelling common factors before solving.
Updated On: Jun 24, 2026
- \(5\)
- \(8\)
- \(10\)
- \(3\)
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The Correct Option is A
Solution and Explanation
Step 1: Write the combinations using formula.
Given, \[ \frac{{}^{2n}C_3}{{}^{n}C_3}=12 \] Using \[ {}^nC_3=\frac{n(n-1)(n-2)}{3!}, \] we get \[ \frac{ \frac{(2n)(2n-1)(2n-2)}{6} }{ \frac{n(n-1)(n-2)}{6} } =12 \] The factor \(6\) cancels: \[ \frac{(2n)(2n-1)(2n-2)}{n(n-1)(n-2)}=12 \]
Step 2: Simplify the expression.
Since \[ 2n-2=2(n-1), \] we get \[ \frac{2n(2n-1)\cdot 2(n-1)}{n(n-1)(n-2)}=12 \] Cancel \(n\) and \((n-1)\): \[ \frac{4(2n-1)}{n-2}=12 \]
Step 3: Solve for \(n\).
\[ 4(2n-1)=12(n-2) \] \[ 8n-4=12n-24 \] \[ 20=4n \] \[ n=5 \]
Step 4: Final conclusion.
Therefore, \[ \boxed{5} \]
Given, \[ \frac{{}^{2n}C_3}{{}^{n}C_3}=12 \] Using \[ {}^nC_3=\frac{n(n-1)(n-2)}{3!}, \] we get \[ \frac{ \frac{(2n)(2n-1)(2n-2)}{6} }{ \frac{n(n-1)(n-2)}{6} } =12 \] The factor \(6\) cancels: \[ \frac{(2n)(2n-1)(2n-2)}{n(n-1)(n-2)}=12 \]
Step 2: Simplify the expression.
Since \[ 2n-2=2(n-1), \] we get \[ \frac{2n(2n-1)\cdot 2(n-1)}{n(n-1)(n-2)}=12 \] Cancel \(n\) and \((n-1)\): \[ \frac{4(2n-1)}{n-2}=12 \]
Step 3: Solve for \(n\).
\[ 4(2n-1)=12(n-2) \] \[ 8n-4=12n-24 \] \[ 20=4n \] \[ n=5 \]
Step 4: Final conclusion.
Therefore, \[ \boxed{5} \]
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