Question 34975: How many ways can you order a hamburger if you can order it with or without the following items: cheese, lettuce, tomatoes, pickles or onions?
Found 3 solutions by stanbon, JWG, richard1234:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! How many ways can you order a hamburger if you can order it with or without the following items: cheese, lettuce, tomatoes, pickles or onions?
You choose to have or you choose to not have. 2 choices
But you do this 5 times
# of orders = 2^5=32 different orders.
Cheers,
Stan H.
Answer by JWG(21) (Show Source):
You can put this solution on YOUR website! My solution is much longer, but same answer.
I read this problem as saying, "How many ways can you make a hamburger with five different fixings?" This means I can make a hamburger with all five fixings, or as little as none--just a plain burger with a bun.
Five Fixings: 1 way
Four Fixings: 5C4 = 5 ways
Three Fixings: 5C3 = 10 ways
Two Fixings: 5C2 = 10 ways
One Fixing: 5 ways
Zero Fixings: 1 way
1+5+10+10+5+1 = 32
32 ways to make a hamburger when there are five fixings to choose from, and amount of the five chosen vary.
Answer by richard1234(7193)  (Show Source):
You can put this solution on YOUR website! Each item can be put on or left on. There are five items, so there are 2^5, or 32, possible combinations of the burger.
Another way to visualize it is to note that all of the binary numbers from 00000 to 11111 represent the numbers 0 through 31, in which there are 32 numbers here.
|
|
|
