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)

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