1.

Determine `n` if (i) `""^(2n)C_(3) : ""^(n)C_(3) = 12:1` (ii) `""^(2n)C_(3): ""^(n)C_(3)= 11:1`

Answer» (i) `C(2n,2):C(n,2) = ((2n!)/((2!)(2n-2)!))/((n!)/((2!)(n-2)!)`
`=(((2n)**(2n-1)**(2n-2)!)/((2!)(2n-2)!))/(((n)**(n-1)**(n-2)!)/((2!)(n-2)!))`
`=(2(2n-1))/(n-1)`
Now, we are given,`C(2n,2):C(n,2) = 12:1`
`:. (2(2n-1))/(n-1) = 12`
`=>4n-2 = 12n-12=> 8n = 10 => n = 5/4`
But, `n` should be a whole number.
So, there is no value of `n` that can solve this.

(ii) `C(2n,3):C(n,3) = ((2n!)/((3!)(2n-3)!))/((n!)/((3!)(n-3)!)`
`=(((2n)**(2n-1)**(2n-2)**(2n-3)!)/((2!)(2n-3)!))/(((n)**(n-1)**(n-2)**(n-3)!)/((3!)(n-3)!))`
`=(2(2n-1)(2n-2))/((n-1)(n-2))`
`=(4(2n-1))/(n-2)`
Now, we are given,`C(2n,3):C(n,3) = 11:1`
`:. (4(2n-1))/(n-2) = 11/1`
`=>8n-4 = 11n-22`
`=> 3n = 18=> n = 6`


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