There are tickets which are numbered from $0$ to $30$. In how many ways can three tickets be withdrawn so that the sum of the numbers is $30$?
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It would help to clarify if there is only one ticket with each number, or if a number can appear on multiple tickets. – hardmath Jul 14 '17 at 10:10
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Hint: Using the stars-and-bars problem, the number of ways to distribute $30$ among three tickets equals:
$${30 + 3 - 1 \choose 3 - 1} = {32 \choose 2} = 496$$
However, this includes cases where two or more numbers are the same. How many invalid pairs of numbers are there? Simply subtract this number from $496$ to arrive at the correct answer.
jvdhooft
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