Question

After a late night of math studying, you and your friends decide to go to your favorite tax-free fast food Mexican restaurant, Burrito Chime. You decide to order off of the dollar menu, which has 7 items. Your group has $16 to spend (and will spend all of it).
a. How many different orders are possible? Explain. (The order in which the order is placed does not matter -just which and how many of each item that is ordered.)
b. How many different orders are possible if you want to get at least one of each item? Explain.
c. How many different orders are possible if you don't get more than 4 of any one item? Explain.

Answer 1

Answer:

Step-by-step explanation:

a. How many different orders are possible?

You have 16 dollars and 7 types of items:

For one dollars, there are different ways to order

For two dollars, there are 7x7 different ways to order

Like wise for 16 dollars there are $7^{16}$ ways of ordering

Answer is = $7^{16}$

b. How many different orders are possible if you want to get at least one of each item?

Now consider that one of each item has to be selected. For first 7 dollars there are 7! ways to do it (Considering the order matters)

If the order doesn't matter there is only one way to do it

Now the remaining 9 dollars, there are $7^{9}$ ways to do it.

If the order matters the answer is 7! × $7^{9}$

If the order doesn't matter the is 1 × $7^{9}$

How many different orders are possible if you don't get more than 4 of any one item?

Now each of the item cannot be ordered than more than 4 times so we have 7×4=28 items to choose from

Now using the permutation formula consider n=28 and r=16

$P=\frac{28!}{(28-16)!}$

$P=\frac{28!}{(12)!}$

Now if the order doesn't matter, use the combination formula

$C=\frac{28!}{12!16!}$