Permutations and Combinations Continued

Five strong contenders are running a race. How many ways can the gold, silver and bronze be awarded? We have five possible athletes to choose from but three at a time. The order does matter here as these are first, second and third places. Also, one person can not be first and second, or repetition is not allowed. So it is a permutation problem.

\\ _nP_r = \frac{n!}{(n-r)!}  \\ _5P_3 = \frac{5!}{(5-3)!} = \frac{5!}{2!} = \frac{5 * 4 * 3 * 2 * 1}{2 * 1} = 60

Now, you have five topping choices to make pizza with three toppings. How many distinct pizzas can you make? The first thing to notice here is the lack of order – your selection of pepperoni, onions and mushrooms is no different from onions, pepperoni and mushrooms or mushrooms, onions, pepperoni etc. It becomes a combination problem.

\\ _nC_r = \frac{n!}{(n-r)! r!} \\ _{5}C_3 = \frac{5!}{(5-3)!3!} = \frac{5!}{2!3!} = \frac{5*4*3*2*1}{(2*1)((3*2*1)} = \frac{5*4}{2} = 10

Not to forget that this problem does not allow you to choose one-topping twice, which real shops may permit!