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If .^36C(r+1) xx (k^2-3) = 6 xx .^35Cr ...

If `.^36C_(r+1) xx (k^2-3) = 6 xx .^35C_r ` , then number of ordered pairs (r, k) are (where `kinI`).

A

4

B

6

C

2

D

3

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The correct Answer is:
`A`
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