Math, asked by abdullahsarkar7714, 1 month ago

There are 5 different balls (Red, Blue, Green, Yellow, Black) that need to be put in 3 different boxes (Box 1, Box 2 & Box 3)



(a) In how many ways can you put 5 different balls in three different boxes? [3 Marks]

(b) In how many ways can you put 5 different balls so that box 1 always gets the green ball? [3 Marks]

(c) In how many ways can you put 5 different balls so that Box 1 gets 2 balls, box 2 gets 2 balls & box 3 gets 1 ball? [3 Marks]

(d) In how many ways can you put 5 different balls so that there is at least 1 ball in each box.​

Answers

Answered by shrekrigaming
0

Step-by-step explanation:

According to the question, we have 5 balls to be placed in 3 boxes where no box remains emptyHence, we can have the following kinds of distribution firstly, where the distribution will be (3,1,1) that is, one box gets three balls and the remaining two boxes get one ball each.

No. of ways to choose the box which gets 3 balls =3 (since 3 boxes are there).

No. of ways to select 3 balls out of 5 is = 5

C

3

.

No. of ways to distribute the rest of the balls =2.

Total no. of ways =60

Secondly, where the distribution will be (1,2,2) that is, one box gets one ball and the remaining two boxes get two balls each.

No. of ways to select 1 balls out of 5 is = 5

C

1

.

No. of ways to choose the box which gets 1 balls =3 (since 3 boxes are there).

No. of ways to select 2 balls out of remaining 4 to go in second box is = 4

C

2

.

Total no. of ways =90

Total:60+90=150.

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