In how many ways can we place 10 identical red balls and 10 identical blue balls into 4 distinct urns if each urn has at least 1 ball? In how many ways can we place  10  identical red balls (R) and  10  identical blue balls (B) into  4  distinct urns if each urn has at least 1 ball?
My attempt: 
Put the minimum number of balls into each urn:
case 1: R R R R (4C0, or 4 choose 0) ways to place 4 red balls, remaining 6R balls can be placed in 4 urns in (6+4-1)C(4-1) or (9 choose 3) ways, and 10B can be placed in (10+4-1)C(4-1) or (13 choose 3) ways, then total ways for this case 4C0*9C3*13C3 = 1*84*286 = 24024
case 2: B R R R, R B R R , etc, total of 4C1 ways to have 1 blue ball and 3 red ones, then for remaining 7R and 9B - 4C1*10C3*12C3 = 4*120*220 = 105600
case 3: B B R R , B R B R etc - following same logic 4C2*11C3*11C3 = 163350
case 4: 4C3*12C3*10C3 = 105600
case 5: 4C4*13C3*9C3 = 24024
Total ways = 24024+105600+163350+105600+24024 = 422598
However, the answer doesn't seem to be correct.  What am I doing wrong?
 A: By the complement rule, the answer = U-L, where:
U is the total number of ways 10 identical red balls (R) and 10 identical blue balls (B) can be placed into 4 distinct urns without constraints
$U={10+4-1 \choose 4-1}^2=81796$
L is number of ways with 1 or more urns having no balls
To find L, let's use inclusion-exclusion principle.  Let set A be the set of all distributions of 10 red balls and 10 blue balls in urns 1,2 and 3.  Set B is the set of all distributions of 10 red balls and 10 blue balls in urns 2,3 and 4; set C is the set of all distributions of 10 red balls and 10 blue balls in urns 1,2 and 4; set D is the set of all distributions of 10 red balls and 10 blue balls in urns 1,3 and 4. Then $L = A\cup B \cup C\cup D$ will be the set of all distributions in which at least one urn is left empty.  By inclusion-exclusion
$L = |A\cup B \cup C\cup D| = |A|+|B|+|C|+|D|-(|A\cap B|+|A\cap C|+|A\cap D|+|B\cap C|+|B\cap D|+|C\cap D|) + (|A \cap B \cap C| + |A \cap B \cap D| + |A \cap C \cap D| + |B \cap C \cap D|) - |A \cap B \cap C \cap D|$
$|A| = |B|=|C|=|D|$ - number of ways 10 identical red balls and 10 identical blue balls can be placed into 3 distinct urns
$|A\cap B|=|A\cap C|=|A\cap D|=|B\cap C|=|B\cap D|=|C\cap D|$ - number of ways 10 identical red balls and 10 identical blue balls can be placed into 2 distinct urns
$|A \cap B \cap C|=|A \cap B \cap D|=|A \cap C \cap D|=|B \cap C \cap D|$ - number of ways 10 identical red balls and 10 identical blue balls can be placed into 1 urn
$|A \cap B \cap C \cap D|$ case when all urns are empty - not possible, therefore zero
$|A| = {10+3-1 \choose 3-1}^2=4356$
$|A\cap B|={10+2-1 \choose 2-1}^2=121$
$|A \cap B \cap C|={10+1-1 \choose 1-1}^2=1$
$L=4356*4 - 121*6 +4=17424-726+4=16702$
The ANSWER = $\boxed{U-L = 81796-16702=65094}$
A: When you do a question with the wording "at least" you almost always approaching it using the $1-$ not $P(A)$ method, because otherwise you are often likely to double count or miscount stuff. 
For example in the way you approached the question, you have over counted in your case two. The reason is that if I put two red balls, with the four urns already having $B R B B$, they are fundamentally the same if I put them in the first urn and then the third urn compared to if I put them in the third urn and then the first urn. It is however different if I put one in he second urn and the other in the third urn. 
This is because the ordering of the urn is taken care of at the step when you made a distinction between $B R B B$ as opposed to $R B B B$, then after that distinction, the three blue urn are no longer distinct. So placing a red ball in the first blue urn is no different to placing in the third blue urn for example. 
Using the other method, you calculate the total number of way of placing the balls, then subtract all the ways with 1 or more urn having no balls. Then you should get the right answer. 
