Question:
If $^{n}C_{2017} = {}^{n}C_{2016}$, then $^{n}C_{4033}$ equals
If $^{n}C_{2017} = {}^{n}C_{2016}$, then $^{n}C_{4033}$ equals
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Whenever consecutive combinations are equal, the index lies exactly at the middle of the row in Pascal's Triangle.
Updated On: Apr 30, 2026
- $1$
- $2016$
- $2017$
- $2033$
- $2019$
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The Correct Option is A
Solution and Explanation
Concept:
Key properties of combinations:
• $^{n}C_r = {}^{n}C_{n-r}$
• If $^{n}C_r = {}^{n}C_{r-1}$, then $r = \frac{n+1}{2}$
Step 1: Use given condition.
\[ ^{n}C_{2017} = {}^{n}C_{2016} \] Using identity: \[ r = \frac{n+1}{2} \] Here, \[ 2017 = \frac{n+1}{2} \]
Step 2: Solve for $n$.
\[ n + 1 = 4034 \] \[ n = 4033 \]
Step 3: Evaluate required value.
\[ ^{n}C_{4033} = {}^{4033}C_{4033} \] \[ = 1 \]
Final Answer: \[ \boxed{1} \]
Key properties of combinations:
• $^{n}C_r = {}^{n}C_{n-r}$
• If $^{n}C_r = {}^{n}C_{r-1}$, then $r = \frac{n+1}{2}$
Step 1: Use given condition.
\[ ^{n}C_{2017} = {}^{n}C_{2016} \] Using identity: \[ r = \frac{n+1}{2} \] Here, \[ 2017 = \frac{n+1}{2} \]
Step 2: Solve for $n$.
\[ n + 1 = 4034 \] \[ n = 4033 \]
Step 3: Evaluate required value.
\[ ^{n}C_{4033} = {}^{4033}C_{4033} \] \[ = 1 \]
Final Answer: \[ \boxed{1} \]
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