Xjyuk

So in order to solve this question, I first found all the number of ways in which every student gets at least one ball, i.e.
$$3^7-3C2 times (2^7-2) - 3C1=1806$$
and then I subtracted it with number of ways where exactly one ball is given to one or two students, i.e.
$$1806-(3times7C1(6C2+6C3+6C4))-(3times7C1times6C1)=1806-1050-126
=630 ways$$
So can anyone tell is this approach is right? and if you are still not able to understand my answer, please ignore it and solve the question. Thanks in advance

We must give three balls to one student and two to the two others. There are $3$ ways to choose which student will get three balls, and then $7choose3$ ways to choose which balls he will get. Distinguish one the remaining students. There are $4choose2$ ways to determine which $2$ of the remaining $4$ balls he will get, giving in all$$37choose34choose2=630$$ ways.

There are $3^7$ ways of distributing the balls, ignoring the restriction. Now how many ways do we have to subtract off because the restriction is violated?

Student A could get zero balls. The remaining balls can then be distributed in $2^7$ ways to students B and C.

Student A could get exactly one ball, in $7$ ways. The remaining balls can then be distributed in $2^6$ ways to students B and C. So far, we need to subtract $2^7 + 7cdot 2^6 = 9cdot 64 = 576$ possibilities. Naively, you could jump ahead and subtract a total of $27cdot 64$ balls, but that would be a mistake.

Student B could get zero balls. The remaining balls can then be distributed in $2^7$ ways to students A and C, but you don't want to double subtract the case where zero balls were distributed to each of A and B, or where A got one ball. So you have to add back in $1$ distribution case for $A$ getting no balls, and $7$ cases where $A$ got one ball.

Student B could get one ball, in $7$ ways. Again, you don't want to double subtract the case where one ball was distributed to each of A and B, or where A got no balls and B got 1.

In the end, you tabulate your answer as follows:

• $3^7 = 2187$ total cases.




• Minus $3$ (who got shorted) times $576$ ways to short a particular student.




• Add back $3$ times $1$ way to give all the balls to one student.




• Add back $6$ times $7 = 42 $ ways to give one ball to one student and no balls to another.




• Add back $3$ times $7cdot 6 = 3 cdot 42 = 126$ ways to give two students one ball apiece.

If there were a way to triply violate the conditions you would have to subtract that back out, but there isn't in this problem.

The total is
$$
2187 - 1728 +3+42+126 = 630
$$
ways to distribute the balls.

Of course, with the insight that the distribution must be some variant on $(3,2,2)$ the problem becomes much less messy, as in the previous answer. But this include/exclude method generalizes without requiring that insight.

There are $binom72$ ways of giving two balls to the first student.

There are $binom52$ ways of giving two balls to the second student.

There are $binom32$ ways of giving two balls to the third student.

Lastly, the last ball can be given to any student, so there are $binom31$ ways to give him.

However, there are repeated cases. For example, the case that we give ball 1&2, 4&5, 6&7 to the first, second, last student respectively and the ball 3 is given to the first student is the same as the case that we give ball 1&3, 4&5, 6&7 to the first, second, last student respectively and the ball 2 is given to the first student. Each case way has appeared $binom32$ times.

Therefore, the answer is $dfracbinom72binom52binom32binom31binom32=boxed630$