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How many kinds of mathematics are out of place?
Let f(n) be the arrangement number of N balls without numbers in the box. Then f (n) = (n-1) * [f (n-1)+f (n-2)]

-Prove-

1 balls are ranked first, and * * * has (n- 1) kinds; -Step 1, following the principle of multiplication.

Row the No.2 ball again, which is divided into two situations-addition principle is used at the back.

Put it in the box 1, and the arrangement of the remaining (n-2) balls is the dislocation arrangement of (n-2) balls, which is f (n-2).

If the box 1 is not put in, the arrangement of (n- 1) balls is the dislocation arrangement of (n- 1) balls, that is, f(n- 1).

So f (n) = (n-1) * [f (n-1)+f (n-2)].

-Calculation-

f( 1)=0

f(2)= 1

f(3)=2*[f(2)+f( 1)]=2

f(4)=3*[f(3)+f(2)]=9

f(5)=4*[f(4)+f(3)]=44

f(6)=5*[f(5)+f(4)]=265