A) \[r=3\]
B) \[r=4\]
C) \[r=6\]
D) \[r=5\]
Correct Answer: B
Solution :
\[\frac{(2n)\ !}{(2n-3)\ !\ .\ 3\ !}\times \frac{2\ !\ \times (n-2)\ !}{n\ !}\ =\frac{44}{3}\] \[\Rightarrow \] \[n\] \[\Rightarrow 4(2n-1)=44\]\[\Rightarrow \]\[2n=12\]\[\Rightarrow \]\[n=6\] Now \[^{6}{{C}_{r}}=15\]\[\Rightarrow \]\[^{6}{{C}_{r}}{{=}^{6}}{{C}_{2}}\] or \[^{6}{{C}_{4}}\]\[\Rightarrow \]\[r=2,\ 4\].
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