Its states that –
If (N+1) pigeons occupy N holes, then some hole must have at least 2 pigeons.
Thus if 5 pigeons occupy 4 holes, then there must be some hole with at least 2 pigeons. It is easy to see why : otherwise, each hole as at most 1 pigeon and the total number of pigeons couldn’t be more than 4.
The Math Behind the Fact:
The pigeonhole principle has many generalizations. For instance:
- If you have N pigeons in K holes, and (N/K) is not an integer, then then some hole must have strictly more than (N/K) pigeons. So 16 pigeons occupying 5 holes means some hole has at least 4 pigeons.
- If you have infinitely many pigeons in finitely many holes, then some hole must have infinitely many pigeons!
- If you have an uncountable number of pigeons in a countable number of holes, then some hole has an uncountable number of pigeons!
The minimum number of cards to be dealt from an arbitrarily shuffled deck of 52 cards to guarantee that three cards are from same suit is
There are 4 sets of cards. So, up till 8 cards there is a chance that no more than 2 cards are from a given set. But, once we pick the 9th one, it should make 3 cards from any one of the sets. So, (C) is the answer.