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
Combination problem
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Last edited by Anindya Madhudor on Sat Sep 07, 2013 6:48 pm, edited 1 time in total.
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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
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
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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
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