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March 22[edit]

Need raw numbers only for article on vehicle registration plates[edit]

I have tried to Google this for days, I keep coming up with answers that give me algebraic formulas and or 1(superscript)36 and things like that. I need raw, countable numbers for laymen like myself whose strong suit is anything but math. Here are the things I need help with and can't seem to find answers for:

how many combinations can be made with 5 numbers and 1 letter, in any combination? =
how many combinations can be made with 6 numbers and 1 letter, in any combination? =
how many combinations can be made with 4 numbers and 2 letters, in any combination? =
how many combinations can be made with 5 numbers and 2 letters, in any combination? =
how many combinations can be made with 3 numbers and 3 letters, in any combination? =
how many combinations can be made with 4 numbers and 3 letters, in any combination? =
how many combinations can be made with 3 numbers and 4 letters, in any combination? =

Thank you in advance for your help!--Kintetsubuffalo (talk) 06:07, 22 March 2018 (UTC)[reply]

There are 10 numbers, and 26 letters, so allowing for repeating letters and numbers, the answer for each of these will be

c=10^{a}\cdot 26^{b}

, where a is the number of numbers, and b is the number of letters. This gives the results, in order: 2,600,000; 26,000,000; 6,760,000; 67,600,000; 17,576,000; 175,760,000; 456,976,000. Iffy★Chat -- 07:42, 22 March 2018 (UTC)[reply]

That answer assumes that the letters vs. numbers must appear in particular locations, which is common on plates in many jurisdictions. For instance 4 numbers and 2 letters where the letters must come first, allowing AB1234 but excluding A12B34. If by "any combination" you wish to count all possible arrangements, then you need to multiply Iffy's results by C(a+b,a). (See Combination.) This gives the final results, in order: 15,600,000; 182,000,000; 101,400,000; 1,419,600,000; 351,520,000; 6,151,600,000; 15,994,160,000. -- ToE 11:20, 22 March 2018 (UTC)[reply]

Thanks, IP69, for catching my paste error. I just fixed it to retain clarity, instead of the more typical striking and underlining. -- ToE 00:09, 23 March 2018 (UTC)[reply]

I edited the above two answers for legibility by using commas for thousands. The last 6,151,600,000 is wrong and should be 15,994,160,000.

A further complication is that if letters and digits can appear in any order, it is dangerous to allow both the letter O and the digit 0 to be used. Even if the license plate uses distinguishable characters for the two, humans transcribing the plate may not know which symbol has which meaning, and the owner of plate EM6F9VO may get in trouble for something the owner of plate EM6F9V0 did. Other letter/number pairs, like I and 1, may be similarly problematic to a lesser degree. If you disallow the digit 0, then the answers are: 9,211,644; 96,722,262; 66,528,540; 838,259,604; 256,258,080; 4,036,064,760; 11,659,742,640. On the other hand, if you disallow the letter O, then they are: 15,000,000; 175,000,000; 93,750,000; 1,312,500,000; 312,500,000; 5,468,750,000; 13,671,875,000.

--69.159.62.113 (talk) 19:34, 22 March 2018 (UTC)[reply]

Thank you all!--Kintetsubuffalo (talk) 02:07, 23 March 2018 (UTC)[reply]

Another probability question[edit]

I believe the handout solution is wrong,[1] but shouldn't my solution come out to the same answer as textbook's? Thank you. 151.202.5.26 (talk) 11:54, 22 March 2018 (UTC)[reply]

The question doesn't have enough information to find