2. There are three black balls numbered with the numbers 1 and 2 and six white balls numbered with the numbers 1, 2, and 3. In how many ways can they be arranged in a row of nine balls, so that among them there are 3 black and 6 white balls?
Solution. The Baraniot number is equal to $\binom{9}{3} 2^{3} 3^{6}=489888$, since the positions where the black balls will be located can be chosen in $\binom{9}{3}$ ways, then on each
of these three places we can place the black ball in two ways, and on each of the remaining 6 places we can place a white ball in 3 ways.