`^nC_r = (n!)/((n-r)!r!)`

`n! = 1xx2xx3xx..........(n-1)xxn`

For example;

`5! = 1xx2xx3xx4xx5`

To do this in easy way we can substitute n-5 as follows.

`n-8 = k`

Then we can say that;

`n-5 = k-8+3 = k+3`

`k! = 1xx2xx3xx.....xxk`

`(k+3)! = 1xx2xx3xx.....xxkxx(k+1)(k+2)(k+3) =k!(k+1)(k+2)(k+3) `

Now let us take the combination.

`^(n-5)C_(n-8)`

`= ^(k+3)C_k`

`= ((k+3)!)/((k+3-k)!(k)`

`= ((k+3)!)/(3!(k!))`

`= (k!(k+1)(k+2)(k+3))/(3!(k!))`

`= ((k+1)(k+2)(k+3))/3!`

Now substitute the value of k with n to get the original answers.

`k+1 = n-8+1 = n-7`

`k+2 = n-6`

`k+3 = n-5`

`3! = 1xx2xx3 = 6`

*So the final answer is;*

`^(n-5)C_(n-8) = ((n-7)(n-6)(n-5))/6`

**Further Reading**

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