Three schools are competing in a cross country race. School A has 6 runners, school B has 5 runners, school C has 4 runners. For scoring purposes, the finishing order of the race only considers the school of each runner. How many different finishing orders are there for the 15 runners?
The number of ways to arrange 15 objects is 15! . This number must be reduced by the number of ways to arrange the 6 A’s, the 5 B’s, and the 4 C’s. (This is the so-called multinomial.)
The answer is: 15! / ( 6! * 5! * 4! ) = 15 * 14 * 13 * 11 * 7 * 3 = 630630
The number of ways to do that is 15!/(4!*5!*6!) = 630,630
References :
The number of ways to arrange 15 objects is 15! . This number must be reduced by the number of ways to arrange the 6 A’s, the 5 B’s, and the 4 C’s. (This is the so-called multinomial.)
The answer is: 15! / ( 6! * 5! * 4! ) = 15 * 14 * 13 * 11 * 7 * 3 = 630630
References :
Good question. Think of it this way:
you have fifteen slots. School A’s six runners go into six of the fifteen slots, perhaps 1,3,5,7,12,15. There are 15C6 ways that you can put the six As into 15 slots. Now you have 15-6=9 slots to pit Bs runners in. There are 9C5 ways to fill the nine slots with five runners. Now the Cs need to go into the remaining slots. Since there are 4 Cs and 4 remaining slots, there is only one way to do that.
So the total number of ways to arrange the As, Bs, and Cs is 15C6*9C5=630630
References :