# Registration of cars

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• Mar 8th 2007, 08:59 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, 09:17 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

answer is AB-1234
is it correct to solve this problem in this way??
• Mar 9th 2007, 05:24 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?

All of your answers assume no repetitions.
• Mar 9th 2007, 08:18 PM
bilalrouf
..
if Repetation is allowed only in alphabet then what will be the answer
• Mar 9th 2007, 11:15 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, 11:18 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