The number of ways in which we get sum 15, when three dice are thrown simultaneously, is

Correct option is B

Given:

Total dice thrown = 3

Required sum = 15

Solution:

Probabilities of getting 15 as the sum from three dice:

1. Case: (5, 5, 5)

This happens in 1 way.

2. Case: (6, 6, 3)

Number of ways: 

$$\frac{3!}{2!} = \frac{3 \times 2 \times 1}{2 \times 1} = 3 \text{ ways.}$$​

Number of ways:

$$3! = 3 \times 2 \times 1 = 6 \text{ ways.}$$​

1 + 3 + 6 = 10  ways.