Registration of cars: number of license plates possible

[bilalrouf]

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

 

[soroban]

Re: Registration of cars

Hello, bilalrouf!

There are 3 ways for the Registration of cars.

\(\displaystyle 1:\;ABC\)-\(\displaystyle 123\;\;\)(3 letters, 3 numbers)

\(\displaystyle 2:\;AB\)-\(\displaystyle 1234\;\;\)(2 letters, 4 numbers)

\(\displaystyle 3:\;A\)-\(\displaystyle 13245\;\;\)(1 letter, 5 numbers)

By which method maximun number of cars can be registered?

\(\displaystyle 1:\;ABC\)-\(\displaystyle 123\)
There are \(\displaystyle 26^3\) choices for the three letters
. . and \(\displaystyle 999\) choices for the 3-digit number.
Total: \(\displaystyle \,26^3\,\times\,999 \:=\:17,558,524\) possible registrations.

\(\displaystyle 2:\;AB\)-\(\displaystyle 1234\)
There are \(\displaystyle 26^2\) choices for the two letters
. . and \(\displaystyle 9,999\) choices for the 4-digit number.
Total: \(\displaystyle \,26^2\,\times 9,999\:=\:6,759,324\) possible registrations.

\(\displaystyle 3:\;A\)-\(\displaystyle 12345\)
There are \(\displaystyle 26\) choices for the one letter
. . and 99,999[/tex] choices for the 5-digit number.
Total: \(\displaystyle \,26\,\times 99,999 \:=\:7,599,974\) possible registrations.


Obviously, plan #1 creates more registrations.

 

[bilalrouf]

..

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??

 

[stapel]

...is it correct to solve this problem in this way??

Since your values do not match those given in the complete worked solution that the tutor provided, no, your method would not appear to be correct.

Eliz.

 

[Denis]

bila, assume only letters A,B,C and digits 1,2 are allowed.
AA : 11,12,21,22
AB : 11,12,21,22
AC : 11,12,21,22
BA : 11,12,21,22
BB : 11,12,21,22
BC : 11,12,21,22
CA : 11,12,21,22
CB : 11,12,21,22
CC : 11,12,21,22

That's 36, or (3^2 * 4).
Does your way = 36?

 

[bilalrouf]

Re: ..

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??

in step 2 (AB-1234) There is mistake in my calculation.

 

[bilalrouf]

correct answer

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 * 5040
= 3276000

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

answer is ABC-123



there was some problem in calculation.i have corrected. Option 1 (ABC-123) is correct answer. the difference between the answer of Soroban and me is that ,in my calculatoin repeation of same digit is not involved.But in both cases answer is same.
:!: