A) p divides N
B) \[{{p}^{2}}\]divides N
C) p cannot divide N
D) none of these
Correct Answer: A
Solution :
[a] \[N{{=}^{2n}}{{C}_{n}}=\frac{(2n)!}{{{(n!)}^{2}}}=\frac{(n+1)(n+2)...(n+n)}{(n!)}\] \[\Rightarrow (n!)N=(n+1)(n+2)...(n+n)\] Since \[n<p<2n,\] so p divides (n+1)(n+2)...(n+n).
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