A) \[3\]
B) \[4\]
C) \[5\]
D) \[8\]
Correct Answer: A
Solution :
We have,\[^{15}{{C}_{3r}}{{=}^{15}}{{C}_{r+3}}\] \[\Rightarrow \] \[^{15}{{C}_{15-3r}}{{=}^{15}}{{C}_{r+3}}\] \[\Rightarrow \] \[15-3r=r+3\] \[\Rightarrow \] \[4r=12\] \[\Rightarrow \] \[r=3\] Note: \[^{n}{{C}_{r}}\] is defined only when \[n,\,\,r>0\] and\[n\ge r\].
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