Balls and Urns!

Group theory will help us here....

The number of ways to divide m+n+p objects into three groups having m,n, and p objects is $\frac{(m+n+p)!}{m! n!p!}$

Two cases must be considered:

• (3,1,1)

Number of ways of forming groups= $\frac{5!}{3!1!1!2!}=10$

Ways of distributing the 3 groups= $10 \times 3!=\boxed{60}$

• (2,2,1)

Number of ways of forming groups= $\frac{5!}{2!2!1!2!}=15$

Ways of distributing the 3 groups= $15 \times 3!=\boxed{90}$

Therefore, total number of ways= $60+90=\boxed{150}$

The total number of ways to distribute 5 distinct balls over 3 urns is $3^5=\boxed{243}$

If we subtract distributions in which one or two urns remain empty, we are done. There are

• $\boxed{3}$ distributions in which two urns remain empty
• $3\cdot (2^5-2) = \boxed{90}$ distributions in which 1 urn remains empty.

The desired number is: $243-90-3 = \boxed{150}$

Let property

P1 be a distribution of the balls such that urn 1 has ≥ 1 ball.

P2 be a distribution of the balls such that urn 2 has ≥ 1 ball.

P3 be a distribution of the balls such that urn 3 has ≥ 1 ball.

Let set

A1 be the set of distributions that do not have P1.

A2 be the set of distributions that do not have P2.

A3 be the set of distributions that do not have P3.

Then, we want $|U| - | A1 \cup A2 \cup A3 | = |U| - | A1| - | A2| - | A3| \\ + | A1\cap A2 | + | A1\cap A3 | + | A2 \cap A3 | \\ - | A1 \cap A2 \cap A3 |.$

$|U|$ = the # of distributions of the 5 distinct balls into the 3 distinct urns = $3^5$

$| A1 | = | A2 | = | A3 | = 2^5$ [The reason for this has been left to the reader as an exercise]

$| A1\cap A2 | = | A1\cap A3 | = | A2 \cap A3 | = 1^5$ [The reason for this has been left to the reader as an exercise]

$| A1 \cap A2 \cap A3 | = 0$ [Not possible, is it?]

Now, plug back the values to the equation above

If the urn are not distinct, there would be $\begin{Bmatrix}5\\3\end{Bmatrix}=25$ ways to do that (see Stirling number of the second kind ).

As the urn are distinct, we can multiply by the permutations of the three urns.

$\begin{Bmatrix}5\\3\end{Bmatrix}·3!=25·6=\boxed{150}$