The number of ways of getting a sum 16 on throwing a dice four times is ________.

Solution

Maths sir · 4 steps

Shift each throw to start at 0, count freely, then remove the cases where a throw goes past 6.

1. Set up the equation. Let the four throws be x₁, x₂, x₃, x₄ with 1 ≤ xᵢ ≤ 6 and x₁ + x₂ + x₃ + x₄ = 16. Put yᵢ = xᵢ - 1. Then 0 ≤ yᵢ ≤ 5 and y₁ + y₂ + y₃ + y₄ = 12.
2. Count without the cap. Non-negative solutions of y₁ + y₂ + y₃ + y₄ = 12 number C(15, 3) = 455. Some of these have a yᵢ ≥ 6, which is not a real dice throw.
3. Remove bad cases. If one chosen yᵢ ≥ 6, subtract 6 from it: the remaining sum is 6, giving C(9, 3) = 84 ways, for each of 4 choices. If two are ≥ 6, the remaining sum is 0, giving C(3, 3) = 1 way, for each of C(4, 2) = 6 pairs. Three cannot all be ≥ 6, since that needs 18.

   N = C(15, 3) - 4C(9, 3) + 6C(3, 3) = 455 - 336 + 6

4. So. The number of ways is 455 - 336 + 6 = 125.

The common slip: 455. Forgetting that a die shows at most 6 and stopping at C(15, 3) gives 455, not 125.

The idea to keep. For dice sums, shift each face to start at 0, use stars and bars, then apply inclusion–exclusion on the faces that cross the maximum. Always add back the double overlaps.

Answer from the NTA final answer key. Solution by PrepWiser.