Jimmy is going on a quick vacation, and is packing some clothes. He has twelve shirts, and wants to take three of them with him. How many ways can Jimmy choose his shirts?
These are the number of ways Jimmy can choose his shirts:
\(\binom{12}{3} = \frac{12!}{9! \cdot 3!} = \frac{12 \cdot 11 \cdot 10}{6} = \boxed{220}\)
+2862 \(\frac{n!}{r!(n-r)!}\) is the formula you always use in these types.
N represents number of items you are choosing out of, and r represents the number of items you are choosing.
+4 Well if he has \(12\) shirts, he has \(12\) to choose from. So \(12\) for the first choice, \(11\) for the second choice (because the first has been used), and \(10\) for the third choice. Thus, he has \(12\cdot11\cdot10\) choices, but they are counted \(6\) times each, so the answer is \(\boxed{220}\).
+220 For this problem, do this...
Use combinations for this problem. Substitute the combination formula to get 12!/9!*3!. Then simplify to get 12*11*10/6 which is 2*11*10 which equals 220.
Jimmy can choose his shirts in 220 ways.
~~Hypotenuisance
