Question:
If
\[
{}^{9}C_3+{}^{9}C_5={}^{10}C_r
\]
for some \(r\in \mathbb{N}\), then \(r=\)
If
\[
{}^{9}C_3+{}^{9}C_5={}^{10}C_r
\]
for some \(r\in \mathbb{N}\), then \(r=\)
Show Hint
Two important identities in combinations are:
\[
{}^nC_r={}^nC_{n-r}
\]
and
\[
{}^nC_r+{}^nC_{r+1}={} ^{n+1}C_{r+1}.
\]
These are frequently used to simplify combination expressions quickly.
Updated On: Jun 24, 2026
- \(3\)
- \(4\)
- \(5\)
- \(7\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Use symmetry property of combinations.
Recall that \[ {}^nC_r={}^nC_{n-r} \] Therefore, \[ {}^9C_5={}^9C_{9-5}={}^9C_4 \] Hence, \[ {}^9C_3+{}^9C_5 = {}^9C_3+{}^9C_4 \]
Step 2: Apply Pascal's identity.
Using the identity \[ {}^nC_r+{}^nC_{r+1}={} ^{n+1}C_{r+1}, \] we get \[ {}^9C_3+{}^9C_4 = {}^{10}C_4 \] Thus, \[ {}^{10}C_r={} ^{10}C_4 \] Hence, \[ r=4 \]
Step 3: Final conclusion.
Therefore, \[ \boxed{4} \]
Recall that \[ {}^nC_r={}^nC_{n-r} \] Therefore, \[ {}^9C_5={}^9C_{9-5}={}^9C_4 \] Hence, \[ {}^9C_3+{}^9C_5 = {}^9C_3+{}^9C_4 \]
Step 2: Apply Pascal's identity.
Using the identity \[ {}^nC_r+{}^nC_{r+1}={} ^{n+1}C_{r+1}, \] we get \[ {}^9C_3+{}^9C_4 = {}^{10}C_4 \] Thus, \[ {}^{10}C_r={} ^{10}C_4 \] Hence, \[ r=4 \]
Step 3: Final conclusion.
Therefore, \[ \boxed{4} \]
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