When three dice are thrown, a minimum sum of 3(1 + 1 + 1) and a maximum sum of 18(6 + 6 + 6) is possible, out of which there are 4 multiples of 4(4, 8, 12 and 16).

Case 1: Sum on the dice is 4

The three ways are: (1, 1, 2); (1, 2, 1) and (2, 1, 1).

Case 2: Sum on the dice is 8

(1 + 1 + 6), this will appear in 3 ways

(1 + 2 + 5), this will appear in 6 ways [3! = 6]

Similarly,

(1 + 3 + 4) in 6 ways, (2 + 2 + 4) in 3 ways and (2 + 3 + 3) in 3 ways.

Case 3: Sum on the dice is 12

(1 + 5 + 6) in 6 ways, (2 + 4 + 6) in 6 ways, (2 + 5 + 5) in 3 ways, (3 + 3 + 6) in 3 ways, (3 + 4 + 5) in 6 ways and (4 + 4 + 4) in 1 way.

Case 4: Sum on the dice is 16

(4 + 6 + 6) in 3 ways and (5 + 5 + 6) in 3 ways.

∴ The total number of ways that anyone scores a point = (3) + (3 + 6 + 6 + 3 + 3) + (6 + 6 + 3 + 3 + 6 + 1) + (3 + 3) = 55 ways

Hence, option 5.