A sack contains 48 marbles of four different colors. Exactly 8 of the marbles are red. Prove that the sack must contain at least 14 marbles which are of the sam

Question

Pigeon hole principle 

Answer 1

Ashok,

 

With 48 marbles total of four different colors and 8 of them known to be red, there are 40 marbles left of unknown color.

 

If at least 6 of these remaining 40 are red, then there are at least 14 red marbles and you have proven the result.

 

If less than six of these are red, then there are at least 35 and no more than 40 marbles of the other three colors.

 

Could you have at least one of the other three colored marbles number at least 14 in number? Here is an example of how this could happen.

 

If the other colors are blue, green, and yellow, then you could have 13 blue and green marbles, which would leave 14 yellow marbles. If there are less than 13 marbles of one or two of the three remaining colors, then the third color would have more than 14 marbles.

 

Another way of explaining this (which is a little cleaner) is this:

 

If one of the other colors applies to at least 14 marbles, then you are done. If not, then the other three colors apply to no more than 13 marbles, for a total of 39 marbles. Since there are 40 marbles left over, this is a contradiction.

 

This should explain the answer. (Pardon me for explaining like a mathematician and using a proof by contradiction, but I do not normally teach or explain this subject.)