Anonymous

derangement, permutation, combination equation

for D drivers driving C cars no one driving their own car

( derangement )

eg. 6 men drive 6 of 10 cars ( 6 = D, 10 = C)

no one driving their own car.

howmany possible driving situations are there?

or arrangements of numerals without any falling in their natural position ( 1,2,3,4,5..... 1 can not be first, 3 cannot be third.... etc. )

this equation works for 3 men 5 cars and 4 men 5 cars and 7 men 10 cars so I am assuming it works for all possible car/driver combinations.

QUESTION: is there a shorthand form of notation for this equation. the patern is obvious ( odd numbers are minus even numbers are plus )

I have no official schooling in this ( self taught ) so maybe this is a well known equation that has a shorthand notation.

any easier way of writing this besides having to litterally write it all out?

Equation:

"total without anyone driving their own car=Answer = A

(Cars_P_Drivers)=(C_P_D)=(X)

A = (C_P_D) - X(D_C_1)/(C_P_1) X(D_C_2)/(C_P_2)

- X(D_C_3)/(C_P_3) X(D_C_4)/(C_P_4).... X(D_C_D)/(C_P_D)

Update:

A = (C_P_D)

- X(D_C_1)/(C_P_1)

+ X(D_C_2)/(C_P_2)

- X(D_C_3)/(C_P_3)

+ X(D_C_4)/(C_P_4)

.... X(D_C_D)/(C_P_D)

Relevance
• Awms A
Lv 7

I don't know any shortcut like the one I gave in the first post on this topic... however, you can simplify the sum a little (at least, I'd call it simplified)

A = D! * ∑ (-1)^n ( (C-n)_C_(C-D) ) / n!

where the sum goes from n = 0 to n = D.

And this actually leads almost directly to why the D! / e shortcut worked in the case of D=C. If D=C, then the combination in the sum equals 1. What we're left with is a Taylor polynomial for e^x evaluated at x = -1 (which is where the division by e came from). Then, since it's an alternating series with terms decreasing in absolute value, the error is no larger than the absolute value of the next term: D! / (D+1)! = 1 / (D+1). D>=1 makes this error < 1/2, so the nearest integer method works.

• 4 years ago