The number of ordered pairs (r k) for which 6.35Cr =(k²-3).36C(r+1) where k is an integer.

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The number of ordered pairs (r, k) for which 6. 35Cr = (k^ – 3). 36Cr + 1, where k is an integer, is The number of ordered pairs (r, k) for which 6. 35Cr = (k^ – 3). 36Cr + 1, where k is an integer, is