IIT JEE , AIEEE Forum - Mathematics , Algebra

shakirshafi12

My answer:
for 2n people
suppose we have arranged the queue and the 5 rupee ones and 10 rupee ones start giving money to the cashier and if the queue stops the number of people having after the queue having 5 rupee coins are irrelevent :
the minimum case required for the line to stop
is that line stops before n-1 people having 5 rupee coins
after that whatever people after are there is irrelevent
therefore
n-1 5ers
n+1 others after
we dont need to know what is thier arrangement cause
this is the least possible case
ways a line can be arranged
=(2n)!/(n!*n!)-(2n)!/((n+1)!*(n-1)!)
divide by total number of ways i.e =(2n)!/(n!*n!)
we will get 1/(n+1)
all guys can you suggest a better solution than
me and rahul

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