Can anyone pls explain this sum?
Ok, so let’s assume A,B,C and D’s orders were Ma,Mb,Mc,Md.
Now All possible ways to arrange the orders will be 4!, correct? What does 4! consist of? It us a sum of
- All wrong orders (4W)
- 1 person get’s the right order, 3 get wrong orders (1R, 3W)
- 2 people get right, 2 wrong (2R,2W)
- All right orders (4R)
*Note that we are not including (3R,1W) because it is not possible. Because if 3 people got the right order, the left over order must also be the right one.
Now, 4! = (4W) + (1R,3W) + (2R,2W) + (4R)
- we don’t know how many ways it can be wrong, so let’s call (4W) = x
- There is only one arrangement that can give you all right orders, so (4R) = 1
- Choose 1 person randomly and give him the right order(4C1). Don’t give the 3 people right orders (2 ways) = 4C1*2
- Choose 2 people randomly and give them right orders (4C2). Don’t give the other 2 right (1 way) = 4C2*1
So simplify for x. Final answer will be x/4!