Discussion
Four fair dices are rolled. Then the number of ways in which the sum of upper faces of four dices can be six, is
Answer: B
The number of ways is 10.
Each die shows at least 1, so write each face as 1 + \(y_{i}\) with \(y_{i} \ge 0\). The sum 6 becomes \(y_{1} + y_{2} + y_{3} + y_{4} = 2\). The number of non-negative solutions is
\[ \binom{2 + 4 - 1}{4 - 1} = \binom{5}{3} = 10 \]
No face exceeds 3 here, so the upper limit of 6 never matters. Checking by listing: (3,1,1,1) in 4 orders and (2,2,1,1) in 6 orders, 10 in all. 4 counts only the first pattern and 6 only the second.
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