We're glad that you asked.

The

**is the***n*th subfactorial__number of permutations of__in which*objects***n**__no object__appears in its natural place (from Wolfram).So, if we have the digits {1, 2, 3, 4} (n=4), we want to know how many ways can we order those numbers such that

**none**appear in their initial position (so, "1" will__not__be first; "2" will__not__be second, "3" will__not__be third, and "4" will__not__be fourth).The following combinations of {1, 2, 3, 4} are the only combinations in which

__no____digit__is in its natural place:{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}

There are

**9 possible combinations**that meet this criterion. So**!4=9**.Remember that there are 4! (=24) possible combinations of {1, 2, 3. 4}, so we note that

**4!-!4=24-9=15**

of them have a value that

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

__do not__meet the criterion (that is,*at least one*of the four digits is in its "natural" place) 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)

[Note that there is no case in which 3 fail, since if 3 of these values are in the

*right*place, the remaining value*must be*in the right place, too.]Mathematically we can use a recursive approach:

**!**

*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

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