Fives in the lottery

Statement

How many fives can be drawn in the lottery, picking 5 numbers out of 90 without replacement and without order?

Solution

Numbered slips (the numbers are distinguishable), without repetition, but the order does not count: these are simple combinations of $90$ objects taken $5$ at a time. $\binom{90}{5} = \frac{90!}{5!\,85!} = \frac{90\cdot 89\cdot 88\cdot 87\cdot 86}{5!} = \boxed{43\,949\,268}$

Links

Topics: Combinatorics
Concepts: Binomial coefficient · Combinations
Methods: Simple combinations
Skills: Combinatorial calculus
Exercise type: Combinatorial problem