find n?

(nC4) = 5(nC3)     we have

n! / [ (n − 4)! 4! ]   =  5  n! / [ (n − 3)! 3! ]    implies that

5 [ [ (n − 4)! 4! ]  =   [ (n − 3)! 3! ]

5 * 4!  * ( n −4)!  = (n − 3)! 3!      rearrange and note that     →    [5 * 4!   = 5!]

5! / 3!   = (n − 3)!  / ( n − 4)!

120/6  = n −  3

20  = n − 3

23  = n

Check

(23C4)   =  8855

5 (23C3)  =  8855