Calculate how many Number-Plates could exist with 3 pairs of numbers or letters

D3V

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I'm not sure if this is the right thread.

I am currently in Portugal and would like to calculate the number of car number plates available.

They are made up as follows.
3 pairs of numbers or letters.
There must be at least one pair of numbers and one pair of letters.

Examples:
- 00 AJ 99
- GH 12 38
- 81 BC 37
- AB CD 44
- 13 37 IT
....

At first I thought it was pretty simple.

102 * 364

But this way the pairs are not taken into account.

Does anyone have an idea what the formula for this would look like ?
 
Consider all the plates with 1 pair of letters and 2 pairs of numbers. The pair of letters can be in one of the 3 positions, and for each of those positions there are [imath]36^4*10^2[/imath] choices. Similarly, for 1 pair of numbers that pair can be in one of the 3 positions, and for each position there are [imath]36^2*10^4[/imath] choices. The total then is [imath]3*36^4*10^2 + 3*36^2*10^4 = 3*36^2*10^2(36^2+10^2) = 542764800[/imath], which is about 51.8 number per capita :)
 
?
That sounds quite reasonable.
But if we take 364, isn't it also
- A1 B2 34
- 56 C7 D8
- E0 99 F1

possible in this formula.

264 would not include the numbers. Or am I making a mistake in thinking.
I guess I should have paid more attention in maths. ?

(PS: i made the calculation...

3 ∗ 364 ∗ 102 + 3 ∗ 362 ∗ 104 = 542'764'800
3 ∗ 264 ∗ 102 + 3 ∗ 262 ∗ 104 = 157'372'800
)

I'm still totally confused...
 
Last edited:
?
That sounds quite reasonable.
But if we take 364, isn't it also
- A1 B2 34
- 56 C7 D8
- E0 99 F1

possible in this formula.

264 would not include the numbers. Or am I making a mistake in thinking.
I guess I should have paid more attention in maths. ?

(PS: i made the calculation...

3 ∗ 364 ∗ 102 + 3 ∗ 362 ∗ 104 = 542'764'800
3 ∗ 264 ∗ 102 + 3 ∗ 262 ∗ 104 = 157'372'800
)

I'm still totally confused...
[imath]36^4[/imath] would not include the numbers but [imath]36^4\times10^2[/imath] would -- does this make sense?
 
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