If 4nC2n2nCn=p×1.3.5.......(4n−1){1.3.5.......(2n−1)}2  then p is

L.H.S.=4n!2n!2n!×n!n!2n!  =(4n)(4n−1)(4n−2)........3.2.1.n.!n!{(2n)(2n−1)(2n−2)......3.2.1.}3  =22n[2n(2n−1)......2.1][1.3.5.....(4n−1)](n!)22n!(2n)2(1.2.3......(n−1)n)2{1.3.5...(2n−1)}2  =1.3.5..(4n−1){1.3.5....(2n−1)}2⇒p=1