In how many ways can a menu be made from 5 dishes, if the menu contains either 3 or 4 dishes?

Correct option is D

Given:

- There are 5 dishes available.
- We need to create a menu containing either 3 or 4 dishes.

We are to find the number of ways to make a menu containing either 3 or 4 dishes.
Solution:

Step 1: Number of ways to choose 3 dishes out of 5
To choose 3 dishes from 5, we use the combination formula:
Combination Formula:$$\binom{n}{r} = \frac{n!}{r!(n - r)!}$$​​

For 3 dishes out of 5:
​$$\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10$$​​

So, there are 10 ways to choose 3 dishes.
Step 2: Number of ways to choose 4 dishes out of 5
To choose 4 dishes from 5:
​$$\binom{5}{4} = \frac{5!}{4!(5-4)!} = \frac{5}{1} = 5$$​​
So, there are 5 ways to choose 4 dishes.
Step 3: Total number of ways
The total number of ways to form the menu is the sum of the two possibilities:
​$$Total \ ways = \binom{5}{3} + \binom{5}{4} = 10 + 5 = 15$$​​
Thus, the total number of ways to create a menu with either 3 or 4 dishes is 15.
Final Answer:
The correct option is (d) 15.