A) 3
B) 6
C) 2
D) 4
Correct Answer: D
Solution :
[d] \[\because \,\,\,\,{{\,}^{36}}{{C}_{r+1}}\cdot ({{k}^{2}}3)\,={{\,}^{35}}{{C}_{r}}\times 6\] \[\frac{36!}{(r+1)!(35-r)!}\cdot ({{k}^{2}}-3)=\frac{35!}{r!(35-r)}\times 6\] \[6({{k}^{2}}-3)=r+1\] \[\therefore \,\,\,\,{{k}^{2}}=3+\frac{r+1}{6}\] \[\therefore \] r can be 5 and 35 When r = 5 then \[k=\pm \,2\]and when r = 35, then\[k=\pm \,3\]. \[\therefore \] Total number of ordered pairs = 4
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