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King Arthur and his eleven honorable knights must sit on a round-table.
In how many ways can you arrange the group, if no honorable knight can
sit between two older honorable knights?
Here are the possible combinations!
Source
What was the challenge?
If king K is sitting at the center then the youngest knight must sit to right or left of the king i.e. 2 possible positions for him.
The second youngest knight now can sit either left or right of the group of 2 made above.
The third youngest knight now can sit either left or right of the group of 3 made above.
And so on.
That is every knight has 2 possible positions except the oldest knight who will have only 1 position left.
This will be make sure each of the knight (except youngest one) has at least 1 younger neighbor (youngest one has king as one neighbor).
So after putting youngest in 2 possible ways the next youngest can be put another 2 possible ways. That is 4 possible combinations for 2.
Similarly, for arranging 3 knights' positions there can be 2^3 = 8 possible combinations.
This way for 10 knights (excluding the oldest which will have only 1 seat available at the end) there are 2^10 = 1024 possible combinations.
