Any math smarties out there? Anyone up for a problem re: lottery numbers?

[Arugula Latte]

My son has been sick and has this problem in his make-up homework (I'm paraphrasing):

Given the numbers 0 through 9, how many different three-digit numbers can be generated?

Thanks! :hi:

[LynzM]

1. Sure, here's an explanation (but not the answer :) )

So you have the digits 0,1,2,3,4,5,6,7,8,9

If you choose a digit, any one of those 10, you have 9 choices left. Then you choose another number and have 8 options left.

So you have 10 choices * 9 choices * 8 choices. An easy way to simplify this type of problem is to scale it down - what if you only had 4 numbers, 0,1,2,3 and were filling two spots, your choices would be:

01
02
03
10
12
13
20
21
23
30
31
32

Or 4 original numbers to choose from and then three numbers to choose from, 4*3.


Wrote the explanation below, first, then realized it's likely not the problem you're actually working on. :rofl:

For three numbers, 0-9 in each position, each position |__| |__| |__| can hold one digit, right?

So you have 0,1,2,3,4,5,6,7,8,9 in each space, or 10 numbers.

If your first number is a 0, your next number can be 0-9, so

00
01
02
03
04
05
06
07
08
09

And for each of those items, your next number could be 0-9 again

010
011
012
013
014
015
016
017
018
019

For example.

So, you have 10 choices, times 10 choices, times 10 more choices. :)

[madinmaryland]

2. Are you a math teacher? That was a really good explanation.

My daughter in fifth grade math had a teacher that taught the so-called "new math". Confusing as hell.

I'm and engineer and took two years of calculus in college and it made no sense whatsoever.

[Chemisse]

14. There is a 'new math' again?

I was boggled by the new (think they called it 'modern') math in 6th grade. We had to learn 'why' for every little thing. And that was 40 years ago.

[LynzM]

17. No, but I tutor math.

I've thought about teaching, as people seem to think I'm good at it. :) But it requires a lot more school...

[HopeHoops]

3. Six, or maybe twenty-seven.

It depends on how much pot you have smoked.

[Ptah]

4. 720

:thumbsup:

[pokerfan]

5. 720 but what concerns me

is that the title said 'lottery numbers.' With lottery numbers, order doesn't matter i.e. 356 = 635.

If order doesn't matter then it's a combination, C(10,3) = 120 which is the ordered answer (720) divided by the number of ways to rearrange three things (6).

But the problem clearly states 'different three digit numbers' so :shrug:

[Dr. Strange]

6. Yeah, I can see several different interpretations here.

If it's three DIFFERENT digits and order doesn't matter, then the answer is 120.
If it's three DIFFERENT digits and order does matter, then the answer is P(10, 3) = 720.
If we're allowed to repeat digits and order matters, then the answer is 10³ = 1000.
If we're allowed to repeat digits and order does not matter, then the answer is C(12, 3) = 220.

Based on this phrasing:
Given the numbers 0 through 9, how many different three-digit numbers can be generated?
I would get 900 as an answer, since the first digit can't be a zero and there is no prohibition on repeating digits. (So, basically, anything from 100 - 999 would be allowed.)

[pokerfan]

7. Ooh, that's a good one!

Following that line of thinking, if you are not allowed to reuse digits but leading zeros are still disallowed (because they don't form three digit numbers) then the answer is 648 (9 * 9 * 8).

Sloppily worded problems are the bane of mathematics.

[Yavin4]

11. "Sloppily worded problems" Is Actually Good Training for Life

As a Project Manager/DBA/Programmer, getting the requirements correct is 99% of the job.

[Yavin4]

10. There's Nothing to Prevent The First Digit from Being Zero

And there's nothing about order.

The one missing piece is whether the numbers can repeat or not.

If they can repeat, then it's 1,000 possibilities.

If they cannot repeat, then it's 720 possibilities.

[pokerfan]

12. Well, it does say right up front that it's a lottery

Order doesn't apply to a lottery.

10 choose 3 = 120

[Arugula Latte]

8. Thanks everyone, for your help!

:hi:

[Yavin4]

9. The Question Is Sort of Incomplete

It depends on whether any of the three digits can repeat or not.


If the numbers can repeat, then


for each digit of the three digits, there are 10 possibilities (0 through 9).


Thus, 10x10x10 = 1,000 possible combinations.




If the numbers cannot repeat, then for the first digit, there are 10 possibilities. For the second digit, there are 9 possibilities, and for the third digit, there are 8 possibilities:

Thus, 10x9x8 = 720.

[Rabrrrrrr]

15. Yep - the question is entirely incomplete. Impossible to answer as it stands.

[Yavin4]

16. Which Begs The Question

Are our kids bad at math or do we not teach it correctly?

[Rabrrrrrr]

13. Think about it - those are the digits we use in numbering.

three digit numbers are 100 to 999.

Ergo, 900 numbers.



However, if what you REALLY meant to ask (and remember, in math problems, CLARITY OF THE PROBLEM IS ESSENTIAL!!!!!!!!!!!!). If this is what you REALLY meant to ask, you were nowhere near it: how many unique three digit combinations can be formed with the numerals 0 through 9, each of which can be used more than once?

Then, the answer is 1000.

OR - if what you meant to ask is - how many unique three digit combinations can be formed with the numerals 0 through 9, each of which can only be used once? (which would be how the powerball part of lottery is done, excepting, obviously, that they use 44 numbers instead of 9), then the answer is 120 (because in this case, order doesn't matter, and so 123 is the same as 321 or 231, as in powerball).

OR - if what you meant to ask is - how many unique three digit NUMBERS can be formed with the numerals 0 through 0, each of which can only be used, and in which order DOES matter (such that 123 IS different from 321), then the answer is 720.

So, to the ludicrously poorly worded question, the answer is 900, 1000, 120, or 720, depending on what you actually mean.


See what a difference it makes when one phrases the question in a non-ambiguous way?

[NNadir]

18. The number of possible permutations of size k taken from n objects is

n!/(n-k)! which others here have called the "combin" or combination function.

n! = n*(n-1)*(n-2)*(n-3)*...*1

Since they are lottery numbers, they are unique.

For 3 numbers selected from ten digits the answer is thus 10!/(10-3)! = 720.