New Year's Countdown Day 21: Present Time!

Santa can distribute the three kinds of toys independently. The number of ways is therefore $a^3$ , where $a$ is the number of ways one type of toy can be distributed.

In principle, seven indistinguishable toys may be divided among seven distinguishable children in $\binom{7+6}6 = 1716$ ways. (A simple star-and-bar argument.)

However, we now have counted several ways in which three or more of the toys go to one child. Consider therefore the situations in which three of the toys are given to one particular child (7 choices here). The remaining 4 toys can be divided in $\binom{4+6}4 = 210$ ways. Thus we subtract $7 \times 210 = 1470$ situations where one child gets too many toys.

But note that we have subtracted too much now; if two children received three or more toys, that situation has been subtracted twice. We make up for this by adding back in the situations where two children ( $\binom 7 2 = 21$ choices) get three of the gift each; the remaining 1 toy can be assigned to 7 children. Thus we add back in $7 \times 21 = 147$ ways.

The solution, then, is $a = 1716 - 1470 + 147 = \boxed{393}.$