Question

Question #64fb5

Answer 1

`n=10`

Explanation:

`""^nC_4 = 21 * ""^(n/2)C_3`

So we have

`(n!)/(4! * (n-4)!) = 21 * ( (n/2)!)/(3! * [(n/2)-3]!`

we have

`(cancel((n-4)!) * n * (n-1) * (n-2) * (n-3))/(4! * cancel([n-4)!]) = 21 * (( cancel((n/2)-4)!) * (n/2) * [(n/2)-1] * [(n/2)-2])/(3! * [cancel((n/2)-4)!])`

`( n * (n-1) * (n-2) * (n-3))/24 = 21 * ( n/2 * [(n/2)-1] * [(n/2) - 2])/6`

`n/2` goes in `n` twice and `((n/2)-1)` goes in `(n-2)` twice

so we get

`( cancel(n) * (n-1) * cancel((n-2)) * (n-3))/24 = 21 * ( cancel(n) * cancel(n-2) * [(n/2) - 2])/(2 * 2 * 6)`

`((n-1) * (n-3))/24 =21 * [(n/2)-2]/24`

This is equivalent to

`(n-1) * (n-3) = 21 * (n/2 - 2)`

Next, we get

`n^2 - 4*n +3 = 21*n/2 - 42`

Bringing everything on one side we get

`n^2 - 14.5*n +45 = 0`

Multiplying the whole thing by `2` we get

`2*n^2 - 29*n +90 = 0`

`2*n^2 - 20*n - 9*n +90 = 0`

`2*n*(n-10) - 9*(n-10) = 0`

`(n-10)*(2*n-9) = 0`

`n=10" "` or `" "n=4.5`

`n` is a natural number, so `n=10` is the valid answer.