$10$ distinct color balls and we want to divide them into $3$ identical boxes with hard restrictions and probability

We have $10$ distinct color balls and we want to divide them into $3$ identical boxes .We decide that two of the groups can contain at most $3$ balls and the rest one contain at most $4$ balls. Moreover, there are two balls called $A$ and $B$ in $10$ balls.

$1-)$ How many ways are there to distribute these $10$ balls to $3$ groups obeying size restriction of groups and $A$ and $B$ are in the same box?

$2-)$ What is the probability that $A$ and $B$ are in the same boxes obeying size restriction of boxes?

$3-)$ How many ways are there to distribute these $10$ balls to $3$ groups obeying size restriction of groups and $A$ and $B$ are not in the same box?

$4-)$ What is the probability that $A$ and $B$ are not in the same boxes obeying size restriction of boxes?

$5-)$ Solve the foregoing $4$ questions if the number of balls is $7$ instead of $10$.

Solution

$1-)$ $$\bigg (\frac{\binom{8}{1}\binom{7}{3}\binom{4}{4}}{2!}=140 \bigg)+ \bigg (\frac{\binom{8}{3}\binom{5}{3}\binom{2}{2}}{2!} =280\bigg )=420$$

$2-)$ $$\bigg (\frac{\binom{8}{1}\binom{7}{3}\binom{4}{4}}{2!}+ \frac{\binom{8}{3}\binom{5}{3}\binom{2}{2}}{2!} \bigg )/\frac{\binom{10}{3}\binom{7}{3}\binom{4}{4}}{2!}=42/168$$

$3-)$ $$\frac{\binom{10}{3}\binom{7}{3}\binom{4}{4}}{2!}-420 =2100-420=1680$$

$4-)$ By using $3$ and $4$ , $$1-42/168=126/168$$

$5-)$ I could not solve this part because of low number of balls, and I do not know how to distribute.

Are my solutions correct? If not, can you correct me and help for $5$?

Solution 1:

First question:

The first term is incorrect. It should be,

$ \displaystyle \binom{8}{1}\binom{7}{3}\binom{4}{4}=280$

Unlike second term, it should not be divided by $2!$ as the boxes having $3$ balls each are different groups - one of them has $A$ and $B$ and the other is without them.

So the answer should be $560$

Second question:

The denominator, as you mentioned, $ \displaystyle \binom{10}{3}\binom{7}{3}\binom{4}{4} / 2! = 2100$

So the probability is $\displaystyle \frac{560}{2100} = \frac{4}{15}$

Now you can use above results to solve $(3)$ and $(4)$.

Coming to the question if there are $7$ balls instead of $10$, you can break it into cases: $i$) three groups with $1, 2, 4$ balls $ii$) three groups with $2, 2, 3$ balls $iii$) $1, 3, 3$ balls