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

Solution:
Given:
We have 5 dishes, and the menu must contain either 3 dishes or 4 dishes.

Concept Used:
The number of ways to select a group of items from a set is determined using combinations, denoted as $$\binom{n}{r}$$​, which counts the ways to choose ( r ) items from ( n ) items.

Formula Used:
​$$\binom{n}{r} = \frac{n!}{r!(n-r)!}$$​​

Solution:
1. Number of ways to choose 3 dishes from 5 dishes:
​$$\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \cdot 4}{2 \cdot 1} = 10$$​​

2. Number of ways to choose 4 dishes from 5 dishes:
​$$\binom{5}{4} = \frac{5!}{4!(5-4)!} = \frac{5}{1} = 5$$​​

3. Total number of ways: $$\text{Total} = \binom{5}{3} + \binom{5}{4} = 10 + 5 = 15$$​​
Answer:
(d) 15