Let s = spots and n = number of students
n! = "n factorial" = product of all numbers n or less
P = n!/s!(n-s)! = 5!/2!(5-2)!
5! = 5 * 4 * 3 * 2 *1 = 120
2! = 2 * 1 = 2
(5-2)! = 3! = 3 * 2 * 1 = 6
P = 120/12 = 10
Jen,amy,linda,max and jenny are compteting to fill two remaining spots on the chess team. How mant different combinations are possible for filling the two spots?
120,60,20,or 10
1 answer
1 answer