The number of ordered pairs r- k for which 6- 35Cr-k2-3- 36Cr-1- where k is an integer- is -

Using ^36C_r+1=36r+1× ^35C_r, we get 36r+1× ^35C_r × (k^2 3)= ^35C_r × 6 ⇒ k^2 3 =r+16 ⇒ k^2=r+16+3 k ∈ I r → Non negative integer 0 ≤ r ≤ 35 r=5 ⇒ k=± 2 r=35 ...

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Standard XI

Mathematics

nCr Definitions and Properties

The number of...

Question

The number of ordered pairs $(r, k)$ for which $6 \cdot^{35} C_{r} = (k^{2} - 3) \cdot^{36} C_{r + 1}$ , where $k$ is an integer, is :

4

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Solution

The correct option is A $4$
Using $^{36} C_{r + 1} = \frac{36}{r + 1} \times^{35} C_{r}$ , we get
$\frac{36}{r + 1} \times^{35} C_{r} \times (k^{2} - 3) =^{35} C_{r} \times 6$
$\Rightarrow k^{2} - 3 = \frac{r + 1}{6}$
$\Rightarrow k^{2} = \frac{r + 1}{6} + 3$
$k \in I$
r $\to$ Non-negative integer $0 \leq r \leq 35$
$r = 5 \Rightarrow k = \pm 2$
$r = 35 \Rightarrow k = \pm 3$
$∴$ No. of ordered pairs $(r, k) = 4$

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The number of ordered pairs (r,k) for which 6⋅35Cr=(k2−3)⋅36Cr+1, where k is an integer, is :

The correct option is A 4Using 36Cr+1=36r+1× 35Cr, we get 36r+1× 35Cr×(k2−3)= 35Cr×6 ⇒k2−3=r+16 ⇒k2=r+16+3 k∈I r → Non-negative integer 0≤r≤35 r=5⇒k=±2 r=35⇒k=±3 ∴ No. of ordered pairs (r,k)=4

The number of ordered pairs $(r, k)$ for which $6 \cdot^{35} C_{r} = (k^{2} - 3) \cdot^{36} C_{r + 1}$ , where $k$ is an integer, is :