# Thread: Wendy's hamburgers! help urgent

1. ## Wendy's hamburgers! help urgent

heres the problem my math teacher gave out. its due 2morrow. 2/29/08

CHEESE, MAYO, KETCHUP, MUSTARD, PICKLES, ONIONS, TOMATOES, LETTUCE. How many ways can you order a hamburger at Wendy's.

8th grade math. I know its easy but how can you do the problem without doing the big tree that takes up a lot of space on your paper.

2. ## :(

come on peoplle i know u are out there.

3. Why yes, there is a way.

Suppose you wanted to order a hamburger with all of those toppings. There is one way to get all 8 toppings. How many ways for two topping?. C(8,2)

Choosing 3 toppings, C(8,3). And so on.

So, we have:

$\sum_{k=0}^{8}C(8,k)$

Can you count them up?. Are you familiar with combinations?.

4. we did combinations in 6th grade but i forget how to them because that was like a year and a half ago......

5. Another option would be to say:

Your burger can be either with cheese or no cheese. Two options.
Your burger can be either with mayo or no mayo. Two options.
Your burger can be either with ketchup or no ketchup. Two options.
And so on through all 8 choices, so the total is:

$2\times 2\times 2\times 2\times 2\times 2\times 2\times 2$, or $2^8$.

6. Originally Posted by Henderson
Another option would be to say:

Your burger can be either with cheese or no cheese. Two options.
Your burger can be either with mayo or no mayo. Two options.
Your burger can be either with ketchup or no ketchup. Two options.
And so on through all 8 choices, so the total is:

$2\times 2\times 2\times 2\times 2\times 2\times 2\times 2$, or $2^8$.
thank you. i will try that.

7. Yes, Henderson beat me. I was going to point out the identity:

$2^{n}=\begin{pmatrix}n\\0\end{pmatrix}+\begin{pmat rix}n\\1\end{pmatrix}+\begin{pmatrix}n\\2\end{pmat rix}+............+\begin{pmatrix}n\\n\end{pmatrix}$

It may come in handy if you have to sum up combinations.

I don't what to get too technical. What grade are you in?. What class is this for?. May I ask?.

8. 7th pre-algebra....
ur not gonna stalk me are you? lol

9. haha, no, I just wanted to get an idea rather I should post the formula or not.

10. if u think i could handle it. and if it helps.

i tried galactus's solution and ended up with 288 ways. I also tried henderson's way and got 256 ways. which is correct? if i did do them correct.

12. It should be 256. $2^8=256$.

With each topping, you can choose to include it or not include it. Thus, you have 2 choices with each topping, so you multiply $2\times 2\times 2\times 2\times 2\times 2\times 2\times 2$.

13. Originally Posted by Mathnasium
It should be 256. $2^8=256$.

With each topping, you can choose to include it or not include it. Thus, you have 2 choices with each topping, so you multiply $2\times 2\times 2\times 2\times 2\times 2\times 2\times 2$.
thank you. i was debating over that for an hour now. hehe

14. You should get 256 with either way. They're the same. But 2^8 is certainly easier than counting up the combinations from 0 to 8. I was just showing the identity.