Binomial Theorem And Its Simple Applications Question 101
Question: p is a prime number and n<p<2n. if N= $ ^{2n}C_{n} $ , then
Options:
A) p divides N
B) $ p^{2} $ divides N
C) p cannot divide N
D) none of these
Show Answer
Answer:
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).
BETA
