First, set three different boxes to ABC.

The title does not say whether it is allowed to be empty, but since it is said that it should be placed in three different boxes, at least one should be considered.

Three white balls =3+6+ 1= 10.

(1) all three in a box A3 B3 C3 = C(3, 1)

② put two in one box, and put one a2b1a2c1b2a1c2a1= p (3,2)

③ put three in a box, a1b1c1= c (3,3).

4 red balls =3+3+6+3= 15.

It's almost the same as the last one, except for an extra ball.

① Put all four in a box A4 B4 C4 = C(3, 1)

② A box with three squares and another box A/KOOC-0/B3 A/KOOC-0/C3B/KOOC-0/C3B/KOOC-0/C3C/KOOC-0/C3C/KOOC-0/B3 = P (3,2)

③ Put two in one box and two in another box. A2B2A2C2B2C2 = C (3,2)

④ Put two boxes in each box, and put two boxes in the other box A 1 b1C2A1B2C1A2b1c1= c (3,1).

Five yellow balls =3+6+6+3+3=2 1.

① All five of them are put in a box, and C(3, 1)= A5 B5 C5.

(2) Four in a box, single P (3,2)

③ Put three in one box and two P (3,2) in another box.

④ Put three in a box and two in C(3, 1).

⑤ Put two in a box and one in C(3, 1).

Because the positions of the three balls are independent of each other, the multiplication rule is10×15× 21= 3150.

It's done.