Six in a row with three siblings together

Question

Six people line up in a row. Three of them are siblings who insist on standing together as a group (in some order, but with no outsider between them). How many arrangements satisfy this?

Accepted Answer

Bundle the 3 siblings into a single block. That leaves 4 units to arrange: the sibling block plus the other 3 people. A row of 4 units gives 4! = 24 orderings. Within the block, the 3 siblings can stand in 3! = 6 internal orders. Multiply: 4! 	imes 3! = 24 	imes 6 = 144. Takeaway. Forcing a group of g people together makes them one block: arrange (n - g + 1) units, then multiply by g! for their internal order. Here 4! 	imes 3! = 144.

Six in a row with three siblings together

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