8 Children are standing in a line outside a ticket counter at zoo. 4 of them have a 1 rupee coin each & the remaining four have a 2 rupee coin each. The entry ticket costs 1 rupee each. If all the arrangements of the 8 children are random, the probability that no child will have to wait for a change, if the cashier at the tickets window has no- change to start with is `K`. Then `15K` is equal to

Answer

3

Explanation

Total arrangements (holding 2 types of coins) =`(8!)/(4!4!)=70`
Let `a_(n)` denote the ways of arrangement `2n` children out of which `n` have one Rs. 1 coin & rest have Rs. 2 coin such that noone waits for change at the couner. Let `x` denote the child having Rs. 1 coins & `y` denote child having Rs. 2 coin in hand.

So, `a_(1)=1`
`a_(2)=2`
`a_(3)=2+1+2=5`
`a_(4)=14`
`:.` required probability `(K)=14/70=1/5implies15K=3`