Please settle this maths argument between me and my dad (perm/combo sorta)

Miscellaneous and Personal Stuff I Must Share
1

Don’t bother telling me we’re freaks, I already know it

My parents are in town for my birthday so my father was driving me to the train station today. Ahead of us we saw a car with the license plate 1USA234. Those aren’t the exact digits but I modified the license plate for fear of some sort of moderator smackdown. Anyway, after some debate about whether it was deliberate or purely coincidental, and the general psychology behind vanity plates, we decided to try to figure out how many possible vanity plates you could create with “USA” and 4 other digits (ETA: because we were trying to figure out the probability of someone getting a plate like that if it was coincidental).

So we both agreed that there are only a limited number of ways the license plate could be configured if you use a 7 character requirement. These individual setups are:

1USA234
USA1234
12USA34
1234USA
Then we both agreed that the perm/combo formulas don’t apply because in number strings like phone numbers or license plates you can reuse each digit and the perm/combo formulas sort all of those out. However, 3 of the 7 places are blocked as a group because we want the block “USA”-leaving 4 possible places for numbers to go.

So we then decided that for each individual set up, there are the following number of total possibilities:

(Total digit possibilities in Position 1) * (Total digit possib. Pos. 2) * (TDP Pos 3) (TDP Pos. 4)

= 10 * 10 * 10 * 10

or 10000 (normally I would write it as an exponent but I don’t know how to do that on a keyboard)

Now, here’s where we differ. My dad thinks that the Ultimate Total is

10000 * 10000 * 10000 * 10000 or 10 raised to 16. He says it could be setup 1 AND setup 2 And setup 3 AND setup 4…hence multiplying.

I think it’s 10000 + 10000 + 10000 + 10000. My rationale is it could be setup 1 OR setup 2 OR setup 3 OR setup 4 and no others…you add the total possibilities of the total number of setups.

We got to the train station and we had to agree to mathbitch tonight because I needed to catch my train.

Seeing as I am the dumbf*ck liberal arts major who hasn’t taken math since high school I’m sure my dad is right but I don’t know why my rationale would be incorrect in this instance.

Could someone weigh in? Or if we’re both wrong, why? My OCD won’t let me let this go.

Thanks!

2

I’d say you were right. There are 10,000 possibilities in each format, so you add the possibilities for each. Each location of the “USA” is independent of the others.

3

I vote for you. Why would you multiply?

4

Don’t forget the following:

1U2S3A4
U1S2A34
12U3S4A

5

I’d also like to add that we were puzzling over whether there was some mathematical way to figure out the total number of setups (by we, I mean “me whilst on the train”). We did it “manually” in this instance by moving USA around but how would you figure that shit out for big number strings where you have one preset block. Inquiring minds want to know.

6

No…USA has to be a block.

7

There are instances in combinatorics that you need to multiply but I don’t think this is one of them.

8

Because your rationale is not incorrect; your father flubbed it.

What? No. Make it easier. Consider plates of this variety: two numerals (call them 5 and 6) and one letter (A). What are the possibilities?

55A
56A
65A
66A

5A5
5A6
6A5
6A6

A55
A56
A65
A66

For a total of twelve. Alternative you could calculate it this way. There are 3 possible locations for the letter A. Given a location, the potential digit sequences of length n from a set of d digits are d^n. Therefore the number of possible letter-and-number sequences, in this case is, 3*d^n.

In your example, there are four possible locations for USA. Each location has 10^4 digit sequences. Accordingly, the total number of USA-digit sequences is 40,000, as you deduced.

9

Ah, I’m going to give him that mini-argument. Thanks.

Therefore I was right in that the probability of a coincidental 1USA234 plate was 1/40000, right?

Also, KimmyGibbler…is there some formula based way of figuring out the total number of setups. For instance, in this case it was just USA and four other numbers, so I wrote out the possibilities on my hand…but what it was a 2 unit block with 7 other digits required to make a 9 character code. Is there some way to figure out all the different possibilities with the blocks without writing them all out and potentially missing one?

10

And there you have it, ladies and gentlemen: Kimmy Gibbler answering questions about combinatorics. *I be done seen about everything . . . *

11

I really hate to be the “nitpick guy” but is there a reason you left out 123USA4? By my count (and by agreeing with others upthread in principle) that gives a grand total of 50000 possibilities of the letter block USA combined with 4 digits on a seven character liscense plate.

12

No, oversight. That’s why I asked if there was a mathematical formula to figure out the total possibilities because I figure that it’s very easy to leave one of them out.

Oh, and actually I just looked down at my hand and that one is on there so I just mistyped it here. But the point still stands-is there a way to figure those possibilities mathematically without resorting to writing them out?

13

PLATE =Tally of combinations
USA0000=1
USA0001=2
USA0002=3
…
USA9998=9,999
USA9999=10,000
0USA000=10,001
0USA001=10,002
…
9USA998=19,999
9USA999=20,000
00USA00=20,001
00USA01=20,002
…
99USA98=29,999
99USA99=30,000
000USA0=30,001
000USA1=30,002
…
999USA8=39,999
999USA9=40,000
0000USA=40,001
0001USA=40,002
…
9998USA=49,999
9999USA=50,000

So the answer is 50,000 combinations or the prob of that plate is 1/50,000. Ask him to name a plate combination that is not in the above list if he thinks the answer is greater than 50,000 combinations.

By the way… there are 5 spots you can put the USA in, not 4. So 50,000 not 40,000. (on preview - I see this has been addressed.)
ETA - I have no idea how to do this mathematically!

14

You’re wrong – you just showed a way to do it mathematically.

15

I don’t think this really anything you’d learn in statistics, but for this example the formula would be: Total Number Of Characters - Number of Characters in Block + 1

So in your case: 7 - 3 + 1 = 5 possibilities
If the block of letters was PUZZLE instead of USA, it would be 7 - 6 + 1 = 2 possibilites (makes sense right?).

Then, once you know the number of block arrangements, you’d multiply that by 10 to the power of the number of unknown digits (7 - Number of Characters in Block).

16

Every way of solving this is mathematical, your way as much as any other.

Anyway, as a general formula, if you have a single word (no matter how long) that you have to keep contiguous among D digits, then the total number of possibilities is 10^D * (D+1) [the number of possibilities for the digit-string without the letters * the number of possibilities for how many digits come before the letters]. In this case, that’s 10^4 * (4 + 1) = 50,000.

17

Thank you all! This has been very helpful and it was great of you all to weigh in and answer my questions. I’ll take this victory for what it is…extremely rare. My dad graduated 3rd in his class at IIT-Bombay and has something like 80 patents. He must not have had enough tea this morning.

:slight_smile:

18

You’re calculating the probability wrong. The probability is the number of ways of getting the USA block divided by the total number of license plates. You calculated the numerator correctly, 50,000. The denominator is number of digit combinations times the number of letter combinations times the number of ways of arranging a block of three letters within 4 digits = 10[sup]4[/sup] x 26[sup]3[/sup] x 5. The probability is then 1 / 26[sup]3[/sup] = 1 / 17,576.

Note that this is exactly the same as the probability of picking “USA” when choosing three letters at random.

19

Oh crap…you’re right. The license plate probability would be 50,000/total number of 7 character license plates possible, right? 1/50,000 is the chance of getting the license plate I saw if you walked in and asked for a vanity plate with the USA block in a 7 character license plate.

We never got as far as the probabilities because we were arguing about the combinations.