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

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Correct option is D

We have 36r+1×Cr 35k2−3=6 Cr 35 ∵Cr+1 n+1=n+1r+1Cr n⇒k2−3=r+16⇒k2=3+r+16⇒r=5, 35For r = 5, k=±2and for r=35,k=±3∴ The number 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:

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