Permutation
For the lunch special, you can choose an entrée, a drink, and one side dish. How many meal choices are there
It depends on how many entrees there are, how many drink choices, and how many side dishes there are to chose from.
yea the other guy is right and after that make a tree table and find out example 3 entrees abc drinks and 789 sides Entree drink side
1 a 7
1 b 7
1 c 7
And so on so try it and it will help!
At a restaurant, you can choose 2 different side
dishes from a list of 10. The number of possible
side dish choices can be?
How many meals can you make when choosing one entrée, one side, and one drink?
To find the number of meal choices, we can use the concept of permutations. A permutation is an arrangement of objects in a specific order.
In this case, we can think of the entrée, drink, and side dish as three different objects that we want to arrange in a specific order.
Since there are three objects to arrange, we can use the formula for permutations of n objects taken r at a time, which is given by:
P(n, r) = n! / (n - r)!
Where n is the total number of objects, and r is the number of objects we want to arrange.
In this case, n = 3 (since we have three choices - entrée, drink, and side dish), and r = 3 (since we want to arrange all three choices).
Plugging in the values, we get:
P(3, 3) = 3! / (3 - 3)!
= 3! / 0!
= 3! / 1
= 3 x 2 x 1 / 1
= 6
Therefore, there are 6 meal choices available for the lunch special.