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 selection1. Mon: Kung pao chicken. 2. Tues: Beef with broccoli3. Wed: Kung pao chicken4. Thurs: Moo shu pork5. Fri: Beef with broccoliA different selection:a. Mon: Beef with broccolib. Tues: Kung pao chickenc. Wed: Kung pao chickend. Thurs: Moo shu porke. Fri: Beef with broccoliNow suppose that in addition to selecting her main course, she also selects between water or tea for her drink. How many ways are there for her to select her lunches?

1) 30240 possible ways to select lunches without considering drinks

2) 967680 possible ways to select lunches without considering drinks

Step-by-step explanation:

Note: we assume that she does not take lunches for the weekend

if she has 10 possible lunches then

- the fist day she can find any of the 10

- the second day she can find 9 different lunches (since one appeared the day before)

- the 3rd she can find 8, the 4th 7 and the 5th 6

this can be summarised as

number of possible arrangements = 10*9*8*7*6 = 10! / (10-5) ! = 30240 possible ways

if she can add tea or water, independently of the food and knowing that she can repeat tea or water for lunch in the week then:

arrangements = food arrangements * drink arrangements = (n! / (n-d) !) * (2^d)

= 10! / (10-5) ! * 2^5 = 30240 * 32 = 967680 possible ways to select her lunches