Factorial (2n) / factorial (n)

how can you simplify this, is it 2^n?? if not can you please show the steps

thanks in advance:)

(2n)!/n! = ?

let us write (2n) ! in expanded form

so,

( 2 n ) ! = 1. 2. 3. 4. 5. 6. ... (2 n - 1 ) . ( 2 n )

= ( 2. 4. 6. ... ( 2 n ) ) • ( 1. 3. 5. ... ( 2 n - 1 ))

= [ 2 ( 1 ). 2 ( 2 ). 2 ( 3 ). ... 2 ( n ) ] • ( 1. 3. 5. ... ( 2 n - 1))

= ( 2. 2. 2. ... n times ) ( 1. 2. 3. ... n ) • ( 1. 3. 5. ... ( 2 n - 1 ))

= ( 2ⁿ ) ( n! ) • ( 1. 3. 5. ... ( 2 n - 1 ))

∴ ( 2 n ) ! = ( 2ⁿ ) ( n! ) • ( 1. 3. 5. ... ( 2 n - 1 ) )

∴ ( 2 n ) ! / n ! = 2 ⁿ ( 1. 3 .5. ... ( 2 n - 1 ) ) .......................

(2n)! / n! =

(2n)(2n - 1)(2n - 2)...(n)(n - 1)(n - 2)...(1)
--------------------------...---------- -----------...-----
(n)(n - 1)(n - 2)...(1)

Thus all numbers from n and below cancel leaving
(2n)! / n! = (2n)(2n - 1)(2n - 2)...(n + 1)

unfortunately the simplest way to express this is actually (2n)! / n! , so we may as well leave it like that.


Added: Interesting work sarp. But it's not exactly a simplification.

(2n)!/n! or 2n!/ n!?
the second one is simply 2, but the first one is more complicated.


2n*2n-1*2n-2*2n-3*2n-4
____________________=
n*n-1*n-2
2*(2n-1)*2*(2n-3) ...and so on since 2n-2 is 2(n-1) and so the n-1 cancels and similarly 2n-4 is 2(n-2) so again n-2 cancels.

2n / n

= 2 ( n ) / n

= 2 n^1-1

= 2 n^0

= 2 ( 1 )

= 2

2n!/n!=2 because we can say n!/n!=1