Respuesta :
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 [tex]7^{16}[/tex] ways of ordering
Answer is = [tex]7^{16}[/tex]
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 [tex]7^{9}[/tex] ways to do it.
If the order matters the answer is 7! × [tex]7^{9}[/tex]
If the order doesn't matter the is 1 × [tex]7^{9}[/tex]
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
[tex]P=\frac{28!}{(28-16)!}[/tex]
[tex]P=\frac{28!}{(12)!}[/tex]
Now if the order doesn't matter, use the combination formula
[tex]C=\frac{28!}{12!16!}[/tex]
