A difficult and tricky problem

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SUMMARY

The problem involves dividing 15 coins, with 5 heads and 10 tails, into two groups such that both groups contain the same number of heads facing up, without flipping any coins. The solution requires a clever manipulation of the coins rather than a straightforward division. By selecting any 5 coins and grouping them, the number of heads in that group will equal the number of heads in the remaining group, thus satisfying the condition. This counterintuitive solution demonstrates the importance of thinking outside conventional methods.

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Homework Statement


You've 15 coins out of which 5 coins are turned with their HEAD facing upwards and the rest 10 have their TAILS facing upwards. Suppose you have been blindfolded. Now you've got to divide these 15 coins in two groups such that each group has the same number of coins have their HEADs facing up. But you cannot overturn any coin. How can you do it?? It may seem practically impossible but it can be really done!
2. The attempt at a solution
I thought it might be a problem of probability. I found the probability of HEAD which is 1/15 i.e. 1/3 which means one out of three. But actually I think it is impossible as to the fact that the number of HEADs is odd and we can't divide it into two equal groups! But in the question it says that it can be done! I suppose there must be some trick in it. Please help me.
 
Last edited:
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agnibho said:

Homework Statement


You've 15 coins out of which 5 coins are turned with their HEAD facing upwards and the rest 15 have their TAILS facing upwards. Suppose you have been blindfolded. Now you've got to divide these 15 coins in two groups such that each group has the same number of coins have their HEADs facing up. But you cannot overturn any coin. How can you do it?? It may seem practically impossible but it can be really done!
2. The attempt at a solution
I thought it might be a problem of probability. I found the probability of HEAD which is 1/15 i.e. 1/3 which means one out of three. But actually I think it is impossible as to the fact that the number of HEADs is odd and we can't divide it into two equal groups! But in the question it says that it can be done! I suppose there must be some trick in it. Please help me.

Sorry to say but one difficulty is that if there is a trick in the question, it could easily be hidden in the less than perfect English used to supply the information here.

Perhaps you divide the coins into two groups, of any size you like, but arrange the coins on edge, to form a couple of "cylinders". That way you have not overturned and coin, just stood all of them on edge??
 
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Suppose you have been blindfolded.

I can't see why it is specified that you be blindfolded.

15 coins? So maybe you cover over one of the heads with a tails-up coin?
 

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