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December 20, 2014

December 20, 2014

Posted by **Mathslover Please help** on Sunday, May 5, 2013 at 11:34pm.

- Math -
**DKS**, Monday, May 6, 2013 at 8:13am289

- Math -
**hans**, Monday, May 6, 2013 at 11:14amcan u explain DKS,please...

- IIT Kharagpur -
**Nagaraju Vemuri**, Wednesday, May 8, 2013 at 1:47pmThe possible sets are 2^5 = 32.

We have one null-set.

We have 5 possible sets with one station stop(1 to 5).

we have 5c2 possible sets with two station stops etc ...

Now we can at max accommodate 9 runners per set.

If we add one more runner, some or other set will have 10 runners.

So (32 * 9) + 1 = 289 is ANSWER.

- Math -
**K**, Saturday, May 11, 2013 at 9:21amWhy the possible sets are 2^5?

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