Combinations of Naviagation Paths

What is the question?

If the right move is represented as R and up move as U then, RRRUU is the one path to reach at B.

UURRR is one more path between points A and B.

URRRU is another way to reach at B.

Further, one can reach at B via RURRU.

So number of such paths are possible.

However, if all paths above are observed, we can conclude that total 5 moves are needed to reach from point A to B. Out of those 5, 3 have to be RIGHT and 2 have to be UP. 

That is, any combination having 3R and 2U in 5 moves will give a valid path to reach at B.

Now, number of ways 3R can be placed in 5 moves can be calculated as - 

C(5,3) = 5!/(5-3)! * 2! = 10.

To sum up, there are 10 paths available between points A and B.
 

Language Barrier in International Meeting

Of the 1985 people attending an international meeting, no one speaks more than five languages, and in any subset of three attendees, at least two speak a common language. Prove that some language is spoken by at least 200 of the attendees.

Here is the proof! 

For The Communication in International Meeting

What was the barrier in the communication?

For any attendee A and B, having no common language there must be C who know the language of either A or B to form a trio as mentioned.

Let's make assumption contradicting the statement made in question. Suppose there are only 198 people who can talk in particular language with A or B. Since A can communicate in 5 languages, there are 5 x 198 = 990 people who can talk with A.

That is 990 people are there who have sharing 1 common language with 1 of 5 languages known by A. Similarly, B also can communicate with 990 more people.

Now, if A and B have no common language then there are only 990 + 990 = 1980 people having potential to become C in the trio. This obviously doesn't cover total of remaining people i.e. 1985 - 2 (A and B) = 1983.

Hence, our assumption goes wrong there. So there must be at least 200 attendees knowing the same language .

"Get Out of The Hell !"

You’re new to hell, and you’re given a choice: You can go directly to the fourth circle, or you can play simultaneous chess games against Alexander Alekhine and Aron Nimzowitsch. Alekhine always plays black and smokes a pipe of brimstone. Nimzowitsch plays white and wears cuff links made of human teeth. Neither has ever lost.

If you can manage even a draw against either player, you’ll be set free. But if they both beat you, you’ll go to the eighth circle for eternity.

What should you do?

This trick will save your life! 

Freedom From The Hell !

How you are trapped in hell?

Alekhine always plays black and Nimzowitsch plays always white. Obviously, you will be forced to choose white against Alekhine and black against Nimzowitsch. 

Wait for Nimzowitsch's first move and then play the same move on Alekhine’s board. Note how Alekhine responds to move & copy that move on Nimzowitsch's board.

This way, effectively Nimzowitsch is playing against Alekhine and you are just transferring moves between to 2 masters.

Since, they never lost a single chess game, there are high chances that the game between two ends in draw.

Even if Alekhine wins with his black then you are winning against Nimzowitsch as you are copying Alekhine's move against Nimzowitsch's white using your black. Same is case if Nimzowitsch wins.

In short, you will end up with either draw with both or win against at least one (against both is impossible). You will be free in any case.

And in fact, you need not to have knowledge of how to play chess to get freedom from the hell.

Probability of The Correct Answer?

This is a popular probability puzzle in which you have to select the correct answer at random from the four options below.

Can you tell, whats the probability of choosing correct answer in this random manner.

1) 1/4
2) 1/2
3) 1
4) 1/4

And the correct answer is......

Finding The Probability of Correct Answer

What was the question?

It can't be 1/4 as 1/4 appears 2 times in given 4 options as probability of correct answer when random selection of option in that case would be 2/4 = 1/2. This will be contradiction.

It can't be 1/2 either since the probability of the 1 correct answer out of 4 available options on random selection would be 1/4. That will be contradiction again.

It can't be 1 too as again probability of the 1 correct answer out of 4 available options on random selection would be 1/4. Once again this is contradiction.

Hence, the probability of the 1 correct answer out of 4 available options on random selection would be 0.