Permutations and combinations doubt

[SoniaKapoor]

Find no of ways of distributing 5 distinct balls to 3 distinct boxes so that no box is empty.

[kangkan]

7 C2..its a formula..i dont know how to derive it :(

[neha 1]

what is the answer ?
is it 237 ?

[SoniaKapoor]

no....ans is 150

[Amit Goyal]

Q. Find number of ways of distributing 5 distinct balls to 3 distinct boxes so that no box is empty.
A.

Number of ways of distributing 5 distinct balls to 3 distinct boxes
= 3 x 3 x 3 x 3 x 3

Number of ways of distributing 5 distinct balls to 2 specific boxes (say, Box 1 and Box 2)
= 2 x 2 x 2 x 2 x 2

Number of ways of distributing 5 distinct balls to 1 specific box (say, Box 1)
= 1 x 1 x 1 x 1 x 1

Number of ways of distributing 5 distinct balls to exactly two boxes so that none of those two box is empty
= 3[(2 x 2 x 2 x 2 x 2) - 2(1 x 1 x 1 x 1 x 1)]

Number of ways of distributing 5 distinct balls to exactly 1 box
= 3[1 x 1 x 1 x 1 x 1]

Number of ways of distributing 5 distinct balls to 3 distinct boxes so that no box is empty
= Number of ways of distributing 5 distinct balls to 3 distinct boxes - Number of ways of distributing 5 distinct balls to exactly two boxes so that none of those two box is empty - Number of ways of distributing 5 distinct balls to exactly 1 box
= [3 x 3 x 3 x 3 x 3] - 3[(2 x 2 x 2 x 2 x 2) - 2(1 x 1 x 1 x 1 x 1)] -  3[1 x 1 x 1 x 1 x 1]
= 243 - 90 - 3
= 150

[sherlock holmes]

Solution to your question may be goes like this.....(5c1 * 4c2 * 2c2 )+ (5c1 * 4c1 * 3c3) + (5c1 * 4c3 * 1c1) +(5c2 * 3c1 * 2c2) + (5c2 * 3c2 * 1c1) +(5c3 * 2c1 * 1c1)

Here "*" represents multiplication and "c" represents combination. for example 5c1 is (5!/(4!*1!)= 5)

now the solution says that we need to find all those ways such that each box consist of atleast 1 ball.
so now if we take 1st case in which first box keeps only a single ball then that can be possible only in 5c1 ways, then if second box also consist of 1 ball the it will be 4c2 and then 3rd box has to have remaining 2 balls and that is possible in 2c2 i.e (5c1 * 4c2 * 2c2 ).
so we take all cases in which first box keeps 1 ball, 2 balls and 3 balls and we will arrange remaining balls in rest of the two boxes such that each box will contain atleast one ball.
= (5.6.1 + 5.4.1 + 5.4.1 + 10.3.1 + 10.2.1)
= 150.

[SINGHAM]

Sonia, where did you get this question?

[SoniaKapoor]

Thank you Sir..
Number of ways of distributing 5 distinct balls to exactly two boxes so that none of those two box is empty
= 3[(2 x 2 x 2 x 2 x 2) - 2(1 x 1 x 1 x 1 x 1)]

I didnt get why you mutiplied by 3 and 2 here...pls explain



@singham...i have a prob assignment..got it fr there

[Amit Goyal]

Number of ways of distributing 5 distinct balls to boxes 1 and 2 so that none of those two box is empty
= Number of ways of distributing 5 distinct balls to boxes 1 and 2
- Number of ways of distributing 5 distinct balls to boxes 1 and 2 such that box 1 is empty
- Number of ways of distributing 5 distinct balls to boxes 1 and 2 such that box 2 is empty
= (2 x 2 x 2 x 2 x 2) - (1 x 1 x 1 x 1 x 1) - (1 x 1 x 1 x 1 x 1)
= [(2 x 2 x 2 x 2 x 2) - 2(1 x 1 x 1 x 1 x 1)]


Number of ways of distributing 5 distinct balls to exactly two boxes so that none of those two box is empty
= Number of ways of distributing 5 distinct balls to boxes 1 and 2 so that none of those two box is empty
+ Number of ways of distributing 5 distinct balls to boxes 2 and 3 so that none of those two box is empty
+ Number of ways of distributing 5 distinct balls to boxes 1 and 3 so that none of those two box is empty
= [(2 x 2 x 2 x 2 x 2) - 2(1 x 1 x 1 x 1 x 1)] + [(2 x 2 x 2 x 2 x 2) - 2(1 x 1 x 1 x 1 x 1)] + [(2 x 2 x 2 x 2 x 2) - 2(1 x 1 x 1 x 1 x 1)]
= 3[(2 x 2 x 2 x 2 x 2) - 2(1 x 1 x 1 x 1 x 1)]