Question:
The least value of $n$ so that ${}^{n-1}C_3 + {}^{n-1}C_4>{}^nC_3$ is
The least value of $n$ so that ${}^{n-1}C_3 + {}^{n-1}C_4>{}^nC_3$ is
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Use Pascal’s identity to transform sum of combinations, then compare binomial values.
Updated On: May 19, 2025
- $11$
- $9$
- $8$
- $7$
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The Correct Option is C
Solution and Explanation
We are given: ${}^{n-1}C_3 + {}^{n-1}C_4>{}^nC_3$
Use identity: ${}^{n}C_r + {}^{n}C_{r+1} = {}^{n+1}C_{r+1}$
So: ${}^{n-1}C_3 + {}^{n-1}C_4 = {}^nC_4$
Inequality becomes: ${}^nC_4>{}^nC_3$
This is true when $n \geq 8$
Check at $n = 8$: ${}^8C_4 = 70>{}^8C_3 = 56$ → holds true
Hence, least such $n$ is $8$
Use identity: ${}^{n}C_r + {}^{n}C_{r+1} = {}^{n+1}C_{r+1}$
So: ${}^{n-1}C_3 + {}^{n-1}C_4 = {}^nC_4$
Inequality becomes: ${}^nC_4>{}^nC_3$
This is true when $n \geq 8$
Check at $n = 8$: ${}^8C_4 = 70>{}^8C_3 = 56$ → holds true
Hence, least such $n$ is $8$
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