Question

A Chinese restaurant offers 10 different lunch specials. Each weekday for one week, Fiona goes to the restaurant and selects a lunch special. How many different ways are there for her to select her lunches for the week? Note that which lunch she orders on which day matters, so the following two selections are considered different. One possible selection: Mon: Kung pao chicken Tues: Beef with broccoli Wed: Kung pao chicken Thurs: Moo shu pork Fri: Beef with broccoli A different selection: Mon: Beef with broccoli Tues: Kung pao chicken Wed: Kung pao chicken Thurs: Moo shu pork Fri: Beef with broccoli

Answer 1

Answer:

$\frac{10!}{(10-5)!}$

Step-by-step explanation:

Fiona will attend the restaurant for 5 days, so she will get to choose only 5 out of the 10 menus. At first, recall that the number of ways in which you can pick k elements out of n, without minding the order is $\binom{n}{k} = \frac{n!}{(n-k)!k!}$.

Now, consider that for each group of k elements that you choose, you can order it in k! different ways. Suppose that you have k boxes, that you want to fill out with the k elements you pick. So, for the first box you have k options, for the second one you have k-1 options. Continuing in this fashion, you will have k! differnt ways of ordering the k elements.

So, in total, when order matters, you have $\binom{n}{k}\cdot k!$ ways of choosing.

For our case, we have n=10 and k=5, which gives us

$\binom{10}{5}\cdot 5! = \frac{10!}{(10-5)! 5!}\cdot 5! = \frac{10!}{(10-5)!}$