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possible combinations for six digit license plates, numbers 0-9 and letters a-z.

Mackenzie3

New member
Feb 23, 2020
1
Hi,

I'm not sure if this is the correct forum so if I need to post elsewhere please let me know.

I'm having trouble with calculating the possible combinations for six digit license plates, numbers 0-9 and letters a-z.

I know the overall combinations are 1,947,792 when repetition is allowed and there are no other requirements, but I'm getting stuck when I have to limit three spots to only numbers and the other three to only letters.

I don't have a fancy calculator. Can someone let me know if this is correct?

10*9*8*26*25*24/6*5*4*3*2*1=3,895,584
 
Last edited:

joypav

Active member
Mar 21, 2017
151
Hi,

I'm not sure if this is the correct forum so if I need to post elsewhere please let me know.

I'm having trouble with calculating the possible combinations for six digit license plates, numbers 0-9 and letters a-z.

I know the overall combinations are 1,947,792 when repetition is allowed and there are no other requirements, but I'm getting stuck when I have to limit three spots to only numbers and the other three to only letters.

I don't have a fancy calculator. Can someone let me know if this is correct?

10*9*8*26*25*24/6*5*4*3*2*1=3,895,584
Is the only difference that three have to be numbers and three have to be letters? Because your calculation "10*9*8*26*25*24" suggests that repetition is not allowed.

However, if repetition is allowed and there must be three letters and three numbers, we can find all unique combinations as follows:

$10 \cdot 10 \cdot 10 = 1000$ number permutations
$26 \cdot 26 \cdot 26 = 17576$ letter permutations

But we're not done yet! Now we need to think about the positioning of the letters/numbers. So we need to see how many different ways the letters/numbers can be ordered.

What we want is to choose three places for, let's say, the letters (then the numbers will automatically take the other three positions). So we can order them in 6 choose 3 different ways.

${6 \choose 3} = \frac{6!}{3!(6-3)!}=20$

Then, finally, there are:

$1000 \cdot 17576 \cdot 20 = 351520000$ unique license plates.