Say there are 5 different coloured ball in each bag (C1,C2, C3, C4, C5)
Now make them color
C1=${\color{Blue} Blue}$
C2=${\color{Green} Green}$
C3=${\color{Yellow} Yellow}$
C4=${\color{Orange} Orange}$
C5=${\color{Red} Red}$
:)
Now, how can we select balls?
Case 1:Say, A can select 1 ball from 5 colors (say C1) from B and other than that color it selects ball from C ( say C2)
Case 2: Say, A can select 1 ball from 5 colors (say C1) from B and same color ball from C
So, what will be selecting possibility for A?
$\binom{5}{1}\times \binom{4}{1}+\binom{5}{1}\times 1$
Now, we should calculate selecting possibility of B
Case 1: Now, B can select any ball from 5 balls(may be same color ball which A chooses) from A, say C1 and other any color say C2 from C
Case 2: B can select any ball from 4 balls(may be same color ball which A chooses) from A, say C1 and again same color ball from C
Case 3: B can select any ball from 4 balls( other color ball which A chooses) from A, say C2 and other color ball from C , say C3
Case 4:B can select any ball from 4 balls( other color ball which A chooses) from A, say C2 and again same color ball from C , say C2
So, selection will be $\left ( \binom{5}{1}\times \binom{4}{1}+\binom{5}{1}\times 1+\binom{4}{1}\times \binom{3}{1}+\binom{4}{1}\times 1\right )$
Now, selecting possibility of C
Case 1: Now, C can select any ball from 5 balls(may be same color ball which A chooses) from A, say C1 and other any color say C2 from B
Case 2: C can select any ball from 4 balls(may be same color ball which A chooses) from A, say C1 and again same color ball from B
Case 3: C can select any ball from 3 balls( other color ball which A and B chooses) from A, say C3 and other color ball from C , say C4
Case 4:C can select any ball from 3 balls( other color ball which A and B chooses) from A, say C3 and again same color ball from C , say C3
$\binom{5}{1}\times \binom{4}{1}+\binom{5}{1}\times 1+\binom{3}{1}\times \binom{2}{1}+\binom{3}{1}\times 1$
Similarly again A chooses
Case 1: Now, A can select any ball from 5 balls(may be same color ball which A chooses) from B, say C1 and other any color say C2 from C
Case 2: A can select any ball from 4 balls(may be same color ball which A chooses) from B, say C1 and again same color ball from C
Case 3: A can select any ball from rest 2 balls( other color ball which A and B and C already chooses) from B, say C4 and other color ball from C , say C5
Case 4:A can select any ball from rest 2 balls( other color ball which A and B and C already chooses) from B, say C4 and again same color ball from C , say C4
Now, what is the selecting possibility
$\binom{5}{1}\times \binom{4}{1}+\binom{5}{1}\times 1+\binom{2}{1}\times 1+\binom{2}{1}\times 1$
Lastly if again B chooses to arrange last 1 ball can select like this
$\left (\binom{5}{1}\times \binom{4}{1}+\binom{5}{1}\times 1+\binom{1}{1}\times 1 \right )$
So, total selection possibility
$\left ( \binom{5}{1}\times \binom{4}{1}+\binom{5}{1}\times 1 \right )$+$\left ( \binom{5}{1}\times \binom{4}{1}+\binom{5}{1}\times 1+\binom{4}{1}\times \binom{3}{1}+\binom{4}{1}\times 1\right )$+$\left ( \binom{5}{1}\times \binom{4}{1}+\binom{5}{1}\times 1+\binom{3}{1}\times \binom{2}{1}+\binom{3}{1}\times 1 \right )$+$\left ( \binom{5}{1}\times \binom{4}{1}+\binom{5}{1}\times 1+\binom{2}{1}\times 1+\binom{2}{1}\times 1 \right )$+$\left (\binom{5}{1}\times \binom{4}{1}+\binom{5}{1}\times 1+\binom{1}{1}\times 1 \right )$
Now, here combination of A,B, C is possible , as A can select first or B can select first or C can select first
in 3!=6 ways
So, no of selection will be
6*($\left ( \binom{5}{1}\times \binom{4}{1}+\binom{5}{1}\times 1 \right )$+$\left ( \binom{5}{1}\times \binom{4}{1}+\binom{5}{1}\times 1+\binom{4}{1}\times \binom{3}{1}+\binom{4}{1}\times 1\right )$+$\left ( \binom{5}{1}\times \binom{4}{1}+\binom{5}{1}\times 1+\binom{3}{1}\times \binom{2}{1}+\binom{3}{1}\times 1 \right )$+$\left ( \binom{5}{1}\times \binom{4}{1}+\binom{5}{1}\times 1+\binom{2}{1}\times 1+\binom{2}{1}\times 1 \right )$+$\left (\binom{5}{1}\times \binom{4}{1}+\binom{5}{1}\times 1+\binom{1}{1}\times 1 \right )$)
