Question

Will give medal!!

Did this and turned it in but got nearly everyone wrong, if you could please explain to me. Thank you :)

1. There are 12 appetizers; 4 are soups, 6 contain meat, and 2 do not. In how
many different orders can 3 different appetizers be brought to the table?
2. In how many different orders can 5 different appetizers of the 12 be brought
to the table?
3. 7. Assume that 3 customers arrive and order different appetizers to share from a
choice of 12 appetizers.
a. Does this problem involve permutations or combinations?
b. Why?
c. In how many ways can this be done?

Answer 1

Each combination of 3 different appetizers can be arranged in 3! permutations.

Answer 2

2. The number of permutations of 12 different appetizers taken without repetitions is given by:\[P(12, 5)=\frac{12!}{(12-5)!}=12\times11\times10\times9\times8=you\ can\ calculate\]

Answer 3

....taken 5 at a time....*

Answer 4

@heyimrick What is the full question "3. 7." ?

Answer 5

- In how many different orders can 5 different appetizers of the 12 be brought
to the table?

This one ^

and

- Assume that 3 customers arrive and order different appetizers to share from a
choice of 12 appetizers.
a. Does this problem involve permutations or combinations?

b. Why?

c. In how many possible ways can this be done?

and this ^^

thank you so much, gave u the medal

@kropot72

Answer 6

oh i didint notice you did - In how many different orders can 5 different appetizers of the 12 be brought to the table?


So i need 1. There are 12 appetizers; 4 are soups, 6 contain meat, and 2 do not. In how many different orders can 3 different appetizers be brought to the table?


and

Assume that 3 customers arrive and order different appetizers to share from a choice of 12 appetizers. a. Does this problem involve permutations or combinations? b. Why? c. In how many possible ways can this be done?

Answer 7

1. Each combination of 3 different appetizers can be arranged in 3! permutations.


3. "Assume that 3 customers arrive and order different appetizers to share from a choice of 12 appetizers. a. Does this problem involve permutations or combinations? b. Why? c. In how many possible ways can this be done? "
The 3 customers are going to share the appetizers. Therefore it can be argued that the order of selection of the 3 appetizers does not matter. In this case combinations are needed. The number of combinations of 12 different appetizers taken 3 at a time is:
\[C(12, 3)=\frac{12!}{3!(12-3)!}=\frac{12\times11\times10}{3\times2}=you\ can\ calculate\]