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Prove that `""^(2"n")"C"_"n" = (2^"n" xx 1 xx 3 xx ... (2"n" - 1))/("n"!)`

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#### Solution

L.H.S = `""^(2"n")"C"_"n"`

= `(2"n"!)/("n"!(2"n" - "n")!) = (2"n"!)/("n"!"n"!)`

= `((2"n")(2"n" - 1)(2"n" - 2)(2"n" - 3) ... 4*3*2*1)/("n"!"n"!)`

Numerator has n tems in wich n tems are even and n tems are odd.

Taking one 2 from the n even terms we get

= `(2("n")(2"n" - 1)(2)("n" - 1)(2"n" - 3) ... 2(2)*3*2(2)*1)/("n"!"n"!)`

= `(2^"n"[("n")("n" - 1)("n" - 2) .... 2*1][(2"n" -1)(2"n" - 3) .....3*1])/("n"!"n"!)`

= `(2^"n" xx "n"! (2"n" - 1(2"n" - 3) .... 3*1))/("n"!"n"!)`

= `(2^"n" xx 1 xx 3 xx 5 ... (2"n" - 3)(2"n" - 1))/("n"!)`

= R.H.S

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