Question

amara needs to create a special tasting menu at her restaurant. She needs to select 4 dishes from 7 available dishes and put them in a tasty sequence. How many unique ways are there to arrange 4 of the 7 dishes?

Answer 1

By taking in mind the number of combinations of 4 out of the 7 dishes, and the possible orders of these 4 dishes, we will find that there are 840 unique ways to arrange 4 of the 7 dishes.

We know that if we have a set of N elements, the total number of different combinations of K elements (K ≤ N) out of the N elements is given by:

$C(N, K) = \frac{N!}{(N - K)!*K!}$

In this particular case, we have:

• N = number of available dishes = 7
• K = number that we need to select = 4

Then the number of different combinations is given by:

$C(7, 4) = \frac{7!}{(7 - 4)!*4!} = \frac{7*6*5}{3*2} = 35$

Now we have 4 dishes selected, but we also need to order them.

• For the first dish, there are 4 options.
• For the second, there are 3 options (as one was already selected).
• For the third, there are 2 options.
• For the last one there is only one option.

The number of different arrangements of the dishes is given by the product between the numbers of options above, this gives:

4*3*2*1 = 24

Then the total number of unique ways of arranging 4 of the 7 dishes is:

C = 24*35 = 840

Where we take in mind the possible number of combinations of 4 out of the 7 dishes, and the possible orders of these 4 dishes.

If you want to learn more, you can read:

https://brainly.com/question/9976085