Registration of cars

• Mar 8th 2007, 07:59 PM
bilalrouf
Registration of cars
there are 3 ways for the Registration of cars.

1: ABC-123 or XYZ-123
2: AB-1234 or YZ-1234
3: A=13245 or Z-12345

by which method maximun number of cars can be registred???

Note: there should be no entry like this .
ABC-000 or AB-0000 or A-00000
• Mar 8th 2007, 08:17 PM
bilalrouf
..
i think we can solve it by finding permutation of each part separately.like this

ABC-123
possible number of Registration ={ permutation of ABC }* {permutation of 123)
= (26 P 3) * (10 P 3)
= 15600 * 720
= 11232000

AB-1234
possible number of registration = (26 P 2) * (10 P 4)
= 650 * 151200
= 98280000

A-12345
possible number of registration = (26 P 1) * 10 P 5
= 26 * 30240
= 786240

is it correct to solve this problem in this way??
• Mar 9th 2007, 04:24 AM
Plato
This problem is in need of clarification.
It is not clear if you are allowing repetition?
Can we have a tag such as ADD1322.
You seem to allow repetition in that you say we cannot have AXY0000.
Or is that letters cannot be repeated but digits can?

• Mar 9th 2007, 07:18 PM
bilalrouf
..
if Repetation is allowed only in alphabet then what will be the answer
• Mar 9th 2007, 10:15 PM
CaptainBlack
Quote:

Originally Posted by bilalrouf

AB-1234
possible number of registration = (26 P 2) * (10 P 4)
= 650 * 151200
= 98280000

A reality check on this shows that this cannot be right. Without any
restriction on the registrations other that 2 alphas followed by 4 numerics
gives 26^2*10^4 possible registrations ~= 6.8 million

RonL
• Mar 9th 2007, 10:18 PM
CaptainBlack
Quote:

Originally Posted by bilalrouf
if Repetation is allowed only in alphabet then what will be the answer

Under almost any restrictions about repetitions AAA-NNN will give more
registrations than the others just because there are so many more symbols
that you can drop into the Alpha positions compared to the Numeric
positions.

RonL