Question:easy

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

The Correct Option is B

Solution and Explanation

Step 1: Apply the symmetry property first.
$^9C_5 = ^9C_{9-5} = ^9C_4$. So the left side becomes $^9C_3 + ^9C_4$.

Step 2: Apply Pascal's identity.
Pascal's identity says $^nC_r + ^nC_{r+1} = ^{n+1}C_{r+1}$. Here: \[ ^9C_3 + ^9C_4 = ^{10}C_4 \]

Step 3: Match with the given equation.
We need $^{10}C_r = ^{10}C_4$. So $r = 4$ is one solution. Since $^{10}C_r = ^{10}C_{10-r}$, another solution is $r = 10-4 = 6$. But since the problem states $r \in \mathbb{N}$ and the options offer single values, $r = 4$ is the primary answer.

Step 4: Verify numerically.
$^9C_3 = 84$, $^9C_5 = 126$. Sum $= 210 = ^{10}C_4$. Correct.

Step 5: Note both identities used.
We used: (i) $^nC_r = ^nC_{n-r}$ and (ii) $^nC_r + ^nC_{r+1} = ^{n+1}C_{r+1}$. These are the two fundamental combination identities.

Step 6: State the answer.
\[ \boxed{r=4} \]
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