The no. of ways in which n distinct balls can be put into n distinct boxes so that exactly one box remains empty...is???
62yes i have made a mistake..
there will be inclusion exclusion here...
we could also try reccursion ..
let me try and give the answer.
11According to me
1) We consider there is only one box with a extra storage. Therefore there will be n! ways to adjust the balls.
2) Now we consider there will be n-1 such boxes.
Therefore the no. of ways will be n!(n-1).
3)Now above we consider only a particular box to be empty . but there are n boxes. Therefore the no. of ways is n!n(n-1).
Pls tell me if i am rite?
11For n = 2 There are 4 ways . It is true practically and by the formula.
Even for n=3, There are 36 ways. Practically and by the formula
62Method:
Chose the cell that will be empty: n ways
Chose the 2 balls that will go together: nC2
Arrange them in n-1 cells: (n-1)!
Cheers [6]
13the actual language in the book is....
The number of ways in which 4 distinct balls can be put into 4 boxes lebelled a,b,c,d so that exactly one box remains empty is....
the ans given is 144
i also got 72...like bhaiyaa...
1n ways of selecting the one empty box...n-1 ways of selecting the box thta has the extra ball..nC2 ways of selecting the two balls to go into the box containing the extra ball...and (n-2)! ways of arranging the rest of the n-2 balls...that makes it n(n-1)!nC2...and F(4)=144...cheers!!