## Trending News

# 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)

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)

### 2 Answers

- Awms ALv 71 decade agoFavorite Answer
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.

- louttitLv 44 years ago
hi, a) If there are no longer any regulations, then 8 human beings can take a seat in any of 8 seats. that's 8nPr* = forty,320 techniques. you additionally can get this answer by way of asserting any of the 8 can take a seat of their determination of seats, the subsequent person chooses from the the rest 6, then the the rest 5,4,3,2 and the eighth person basically has a million determination of the final seat. by way of the classic counting theory, you would be able to additionally multiply 8 x 7 x 6 x 5 x 4 x 3 x 2 x a million = forty,320. b) If couples will take a seat mutually each and each couple can take a seat 2 techniques, male-lady or lady-male. yet then the 4 couples can take a seat in diverse arrangements, the place the 1st pair of spots could be taken by way of any of four couples, the subsequent pair has a sort of any of the three remaing couples' seats, then basically 2 options and the final couple takes what's left. that's often completed by way of 4 x 3 x 2 x a million or 4nPr$, the two of which aspects the association of couples can ensue 24 techniques. Then keep in mind each and each couple can take a seat in 2 diverse techniques (mf or fm) so there truthfully are 2 x 24 or forty 8 diverse techniques they are in a position to be seated, protecting the couples mutually. c) 4 women mutually can take a seat 4 x 3 x 2 x a million 0r 24 techniques. (that's comparable to 4 nPr4 the adult adult males could additionally take a seat in 24 diverse arrangements between themselves. This time the communities won't be in a position to turn-flop because of the fact the adult adult males ought to take a seat to the the perfect option of the ladies. So by way of the classic counting theory, there are 24 x 24 or 576 techniques they are in a position to be seated. i'm hoping that enables!! :-)