Question 1203989: A lottery exists where balls numbered 1 to 18 are placed in an urn. Five balls are randomly selected. To win you must match all five balls in any order. How many possible outcomes are there for this game?
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3816)   (Show Source):
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Answer: 8568
Work Shown:
n = 18 balls total
r = 5 selections
Order doesn't matter, so we use the nCr combination formula.
n C r = (n!)/(r!(n-r)!)
18 C 5 = (18!)/(5!*(18-5)!)
18 C 5 = (18!)/(5!*13!)
18 C 5 = (18*17*16*15*14*13!)/(5!*13!)
18 C 5 = (18*17*16*15*14)/(5!)
18 C 5 = (18*17*16*15*14)/(5*4*3*2*1)
18 C 5 = 1028160/120
18 C 5 = 8568
Another approach:
There are 18*17*16*15*14 = 1028160 permutations possible. Start at 18 and count down until we fill 5 slots.
Divide that permutation by 5! = 5*4*3*2*1 = 120 to correct for over-counting.
For instance, the set {1,2,3,4,5} is the same as {1,3,2,4,5} since order doesn't matter. There are 120 ways to arrange any set of five items.
This is how we arrive at 1028160/120 = 8568 different ways to play the game.
Answer by ikleyn(52770)  (Show Source):
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