Math problem: 7 people in a row. A, B and C are arranged from left to right, but not necessarily adjacent. How many arrangements are there? Thank you.

I think it's 7! /3! =840

If you don't consider the order of a, b and c, then it is seven! Seed metering method, but now considering that the order of a, b and c must be divided by 3! Or you can consider 7 years! Regardless of the order, the order of A, B and C can be A, C, B, B, C, B and C (that is, 3! )

Since Party A, Party B and Party C have the same positioning, the probability of six possibilities is equal, so 7! Divided by 6 (that is, 3! You can get all six situations.

So there are 7 people in a row, A, B and C are arranged from left to right, but they are not necessarily adjacent to 7. /3! =840

Species arrangement method

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