At least one is not needed now, so it can be 0.
(1) Putting 0 balls in a box is equivalent to "putting 7 balls in 3 boxes with at least one ball in each box", and there are: C6(2)*C4( 1)=60 kinds.
(2) There are two boxes with zeros, which is equivalent to: "There are seven balls in the two boxes. . . . There are: C6( 1)*C4(2)=36.
(3) If there are three boxes containing 0 balls, that is, 7 balls are put in one box, then there are: C4(3)=4 kinds.
Therefore, * * * has: 20+60+36+4= 120 species.
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This method online is also good:
All 7 balls are put in 4 boxes, and there can be 0 balls.
If you put one in each box first, it is:
Put all 7+4= 1 1 balls into 4 boxes, each box has at least 1 balls. Use the baffle method:
1 1 There is a gap of 1 10 between balls. By inserting three baffles, all the balls of 1 1 can be divided into four boxes.
There are C3/ 10= 120 exposures.
Similarly, put n balls in m boxes,
That is, (n+m) balls are all put into m boxes, and each box has at least 1 ball.
There are C(m- 1, n+m- 1) methods.
All 7 balls are put in 4 boxes, and there can be 0 balls.
Yes:
7+4= 1 1 balls are all put into 4 boxes, and each box has at least 1 balls. Use the baffle method:
1 1 There is a gap of 1 10 between balls. By inserting three baffles, all the balls of 1 1 can be divided into four boxes.
There are C3/ 10= 120 exposures.
Similarly, put n balls in m boxes,
That is, (n+m) balls are all put into m boxes, and each box has at least 1 ball.
There are C(m- 1, n+m- 1) methods.