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The

**is the***n*th subfactorial__number of permutations of__in which*objects***n**__no object__appears in its natural place (from Wolfram).### Example

If we have the digits {1, 2, 3, 4} (n=4), the result is:

!

*n*=(*n*-1)[!(*n*-2)+!(*n*-1)] with !0=1 and !1=0Then

!4=(3)[!2+!3] so we need !2 and !3

!3=(2)[!1+!2]=2[0+1]=2 so we still need !2

!2=(1)[!0+!1]=1[1+0]=1

Finally,

!4=(3)[!2+!3]=3[1+2]=9

Let's check that out.

The following combinations are the only ones in which

__no__digit is in its natural place (1 is not in the first location, 2 is not in the second, 3 is not in the third, 4 in the fourth):2, 1, 4, 3

2, 3, 4, 1

2, 4, 1, 3

3, 1, 4, 2

3, 4, 1, 2

3, 4, 2, 1

4, 1, 2, 3

4, 3, 1, 2

4, 3, 2, 1

As we calculated, there are

**9 possible combinations**. No surprise!But there are 4! (=24) possible combinations of 1; 2; 3; 4 and n!-!n=24-9=15 of them have a value that

*is*in its natural place.The following list is those combinations that

__do not__meet the requirement along with an explanation of why these combinations do not meet the requirement:1, 2, 3, 4 (all digits fail)

1, 2, 4, 3 (1 and 2 fail)

1, 3, 2, 4 (1 and 4 fail)

1, 3, 4, 2 (1 fails)

1, 4, 2, 3 (1 fails)

1, 4, 3, 2 (1 and 3 fail)

2, 1, 3, 4 (3 and 4 fail)

2, 3, 1, 4 (4 fails)

2, 4, 3, 1 (3 fails)

3, 1, 2, 4 (4 fails)

3, 2, 1, 4 (2 and 4 fail)

3, 2, 4, 1 (2 fails)

4, 1, 3, 2 (3 fails)

4, 2, 1, 3 (2 fails)

4, 2, 3, 1 (2 and 3 fail)

On a scale from !0 to !4, how happy are you that you asked?