Stars and Bars Prob #1

(Ignore the pluralisation of dice in this solution, I know it's probably wrong, but it's easier for people to understand). We all know that each dice must have at least 1 dot, meaning we can give each dice 1 dot, leaving us with 3 more dots to apply. We need to decide how many ways we can split up these 3 dots. We could give a single dot to 3 different dice, giving us $\binom{7}{3} = 35$ ways. We could give 2 dots to a dice and 1 dot to a difference dice, giving us $\binom{7}{2} = 21$ ways, however, we multiply this by 2, seen as it matters which of the two dice chosen has the 2 dots, and which has the 1 dot. This gives us $21 \cdot 2 = 42$ ways. Lastly, we could give all 3 dots to a single dice, in $\binom{7}{1} = 7$ ways. Therefore we have $35 + 42 + 7 = \boxed{84}$ possible ways.

Just do stars and bars, and (3+7-1)C(3)

Rolling 7 dice for a sum of 10 can only be done in 3 different ways.

Case 1: Four 1s and Three 2s. Because each die of the same number is indistinguishable, we need to account for the over counting. Thus, this case can be ordered in $\frac{7!}{4! * 3!} = 35$ ways.

Case 2: One 3, one 2, and five 1s. By similar logic, this case accounts for $\frac{7!}{5! * 1! * 1!} = 42$ ways.

Case 3: One 4 and six 1s. By similar logic, this case accounts for $\frac{7!}{6! * 1!} = 7$ ways.

Adding up the separate cases and we get $35 + 42 + 7 = \boxed{84}$ ways.