Question

Four teachers decide to swap desks at work. How many ways can this be done if noteacher is to sit at their previous desk?

Answer 1

To determine the numbers of ways this can be done we can draw a tree diagram. Let T1, T2, T3 and T4 be teachers one, two, three and four. Let A, B, C and D be their original desks.

To teacher one we can assign desk B, C or D. Let's say we assign him desk B, then to the following teacher we can assign desk C or D, if we assign it C the last one will get D automatically. This is representd by the first branch of the tree diagram shown below:

Now, following the same procedure as before we have a diagram for all possibilities starting with teacher one (excluding assign him A since this can't be done) If we start from the first column the branches will tell us which desk is assign to which teacher, for example, the second branch means:

T1-B

T2-A

T3- D

T4-C

In thi diagram we included all possibilities for the other teachers we but have to exclude those possibilities that repeat the desk to a teacher; for example we have to exclude all arrangements that assign to the second teacher the desk B since this is his original desk. Excluding them we notice that of all the posibilities in the tree diagram only nine are viable given the conditions.

Therefore the total number of ways the swap can be done is 9.

This problem can be solved by the derangement formula given by:

`D_n=n!\sum ^n_(k\mathop=0)((-1)^k)/(k!)`

In this case we have four desk to swap, then we have:

`\begin{gathered} D_4=4!\sum ^4_{k\mathop{=}0}((-1)^k)/(k!) \\ =4!\lbrack((-1)^0)/(0!)+((-1)^1)/(1!)+((-1)^2)/(2!)+((-1)^3)/(3!)+((-1)^4)/(4!)\rbrack \\ =4!\lbrack\frac{1^{}}{0!}-(1)/(1!)+\frac{1^{}}{2!}-\frac{1^{}}{3!}+(1)/(4!)\rbrack \\ =24((1)/(1)-(1)/(1)+(1)/(2)-(1)/(6)+(1)/(24)) \\ =24((3)/(8)) \\ =9 \end{gathered}`