Alex's favorite restaurant is a buffet-style Chinese restaurant downtown. Each time he goes, he chooses 2 appetizers and 1 entree. If the restaurant has four appetizers and 5 entrees to choose from, how many, times could Alex go to the restaurant without choosing the same combination of dishes?
I know how to do this question. My question is : Is this a permutation or combination? If you can you please solve this by using the formula like nPr or nCr?
I would consider it a "combination" , assuming that the order that you choose the appetizers should not matter
Number of ways
= C(4,2) x C(5,1)
= 6x5
= 30
To determine whether this problem involves permutations or combinations, we need to consider whether the order of the chosen dishes matters.
In this case, since the order of the dishes does not matter, it is a combination problem. We want to find the number of ways we can choose 2 appetizers out of 4 and 1 entree out of 5.
To solve this using the combination formula nCr, where n represents the total number of items to choose from and r represents the number of items to be chosen, we have:
Number of ways to choose the appetizers: 4C2 = 4! / (2! * (4-2)!) = 6
Number of ways to choose the entree: 5C1 = 5! / (1! * (5-1)!) = 5
Since the choices of appetizers and entrees are independent, we can multiply the number of ways for each to find the total number of different combinations:
Total number of different combinations = 6 * 5 = 30.
Therefore, Alex could go to the restaurant without choosing the same combination of dishes 30 different times.