Permutations And Combinations

Question

# An odd number of stones lie along a straight path, the distance between any two consecutive stones being 10 m. The stones are to the collected at the place where the middle stone lies. A man can carry only one stone at a time. He starts carrying the stones beginning from the extreme stone. If he covers a path of 3 km, how many stones are there ?

Solution

Let the total number of stones be 2n + 1
∴  There are n stones on each side of the middle stone. Let the man starts collecting stones from the extreme left stone.  The distance covered by man to bring the extreme left stone to the middle stone = 10n metres. The distance covered by man to bring the (n - 1)th stone on left to the middle stone = 2 10 (n - 1) metres.
∴   The total distance covered by the man to bring all the stones from left to the middle stone
= 10n + 20(n - 1) + 20(n - 2) + ......... + 20
Similarly, the total distance covered by the man to bring all the stones from the right to the middle stone
= 20 + 20 (n - 1) + ......... + 20
∴     Total distance covered by the man to bring all the stones to the middle stone
= 30n + 40 (n - 1) + 40 (n - 2) +........ + 40 (1)
= 40 [n + (n - 1) + (n - 2) + ....... + 1] - 10n =

But the distance covered is 3 km or 3000 metres.
∴

n = 12
∴      Number of stones = 2(12) + 1 = 25