Combination problem

[Anindya Madhudor]

If n balls are placed at random into n cells, find the probability that exactly one cell remains empty.

A. n/2n!
B. n!/n^n
C. (2n)!/n^n
D. (nC2)* n!/n^n
E. (nC2)* (2n)!/n^n

OA: D

[sanjoy18]

lets try with n=1 that means 1 ball with 1 cells

Then no cells will remain empty.means this "exactly one cell remains empty" is an impossible event
therefore Probability would be 0

putting n=1 in option

A) 2 (probability >1)..hence out
b) 1/1=1 out again
c) 2/1=2 out again
d) we can't put n=1 here as 1c2 not defined..need to further testing
e) same as d

now test =2..

total number exhaustive cases would be 2^2=4
favorable case 1) first cell empty and two balls gone to 2nd cells
2) two balls gone to first cell and 2nd cell is empty

hence probability would be 2/2^2= 1/2

put n=2 in D and E

D) 2/2^2=1/2
E) 4!/2^2=24/4=6 probability greater than 1..out

Hence the answer D

note: we don't need to calculate the probability for n=2 as for n=2 option E's value is greater than 1 hence wrong..Only option satisfy this is d hence right answer

[sanjoy18]

If we want to solve it in general way

The number of ways to place n balls in n cells =n^n.

There are n ways to specify the empty cell. There are n-1 ways of choosing the cell
with two balls. There are nc2 ways of picking the 2 balls to go into this cell. And there are
(n-2)! ways of placing the remaining n-2 balls into the n-2 cells, one ball in each cell.
Hence number of ways of placing the balls such that exactly one cell is
empty is n*(n-1)*(n-2)! *nc2= n!*nc2

Therefore the probability is n!*nc2/n^n

[Java_85]

kinda tough question, I chose D after 10 mins