Figure out this Math Question for Me...

My aunt just texted me...

"I have a 6th grade math question and was wondering if you could anser it:

how many four-digit non-repeating number combos can you make?

If you know the answer can you tell me how you figured it out? Thank you doll! <3"

Can anyone help me out? Is there an easy way to figure this out, like a formula or something?

Try it with a simpler problem first.
The combination function in excel you just put = combin(4,2) and you get the answer 6. That tells you there are six ways to choose 2 things out of 4.
From ABCD you get { AB AC AD BC BD CD } which are all six combinations that are non-repeating (means AB and BA are one answer).


Your question is =combin(10,4) which equals 210 since you are choosing 4 digits out of 10. It's too many to write by hand.

If your aunt means 4567 and 7654 to be different, that's a different question and the answer is much larger than 210 (it is actually 5040). I don't think that is what she means, but you should check.

If you don't have Excel or another spreadsheet, you must do the problem by hand, using the factorial function N!, and a lot of cancelling
combin(10,4)=10!/ (4!*(10-4)! ) = 10*9*8*7*6*5*4*3*2*1 / (4*3*2*1 * 6*5*4*3*2*1 ) = 10*9*8*7 / (4*3*2*1 ) = 10 * 3 * 7 = 210

It's not 6th grade mathematics. It's at least high school, and may be a minor part of many curriculums.

If it actually is supposed to be a 6th grade problem, then it probably really is the permutations version rather than combinations. And the trick may be that the first digit cannot be zero for it to still be a common, 4-digit number. If that is the problem being asked, then the 6th grader would look at it as 9 possibilities for the first digit, 9 for the second, 8 for the third, and 7 for the fourth, giving 9*9*8*7=4,536.

Still really need to clarify what the question really is.

I could be wrong but I'm guessing that 1234 and 4321 are ok. I'm guessing she means just not repeating in each 4-digit number combo.

Quote: HotBlonde

I could be wrong but I'm guessing that 1234 and 4321 are ok. I'm guessing she means just not repeating in each 4-digit number combo.

Usually when someone says "non repeating combination" they mean that 1308 and 8031 are only to be counted once. The technical term for 1308 and 8031 is that they are permutations of one another.

Quote: pacomartin

Usually when someone says "non repeating combination" they mean that 1308 and 8031 are only to be counted once. The technical term for 1308 and 8031 is that they are permutations of one another.

In my previous post, I was interpreting the question as, "How many of the numbers from 1,000 through 9,999 do not have any repeated digits?"

Hi,

Having a math degree I should be able to answer this. Unfortunately the question is ambiguous and therefore the answer depends on what you mean by "non-repeating combos". Furthermore, are there any restrictions on the first digit or any digit at all (one post excluded 0000-0999). One interpretation that nobody has considered is that in a 4 digit number, no digit can be the same as the digits around it. Thus 1032 & 3201 are distinct, whereas 1108 and 3664 are not allowed. But let's also consider the first part of the question, something a 6th grader can do! I would argue that no 6 grader would write "Nine Hundred and Forty Two" as 0942, but 942 so 0000-0999 are out. A 6 grader has never even herd of combinatorics, let alone nCr or nPr or any other statistical notions. A 6 grader does know how to multiply though, and may fathom that the digits in this number are not related to each other (mutually exclusive events?!). In terms of probability, 6 graders tend to be very weak in this subject (my sister actually teaches 7th grade math and told me that they don't even cover probability to there "college" bound students till 8th grade because its too "abstract"). I would argue that the limit of their computational abilities in probability would be to determine whats the probability of a particular sequence of the nightly 4-digit to come up (1/10000 !!!).

Back to the answer: so our first digit can take on 9 values (1-9). The second digit has 10 possible values (0-9), but one of them would repeat the previous digit, so 9 here as well. But guess what, the same is true for the 3rd digit and the 4th (and for that matter any digit even if the problem were extended to an n-digit problem). Thus:

non-repeating # = a1x10^3 + a2x10^2 + a3x10^1 + a4x10^0 with a1[1,9], a2[0-9]/a1, a3[0-9]/a2, a4[0-9]/a3 -> a1=a2=a3=a4=...=an=an+1=....=9

# of comb = 9^n -> #=6561 (if n=4)

I am certain that there are many other interpretations of this question. HB, think of one and I'm sure I can answer it...

jet

Wow, that's a 6th grade question? Just look at the responses above!

In the series "Are you smarter than a 5th grader" only 2 people have ever won it, one of which being a Nobel laureate in physics (A cosmologist), and the other a superintendent of the public school system for an entire state. I would argue that just because a question in principal can be answer by a 6 grader, that that same question couldn't be answered by every adult...

This is an old lottery ticket question...

Considering there are 24 ways to re-arrange the four digits in the form ABCD (10*9*8*7), one would need to fill out 210 tickets to cover each family of 24. This wager would cover 5,040 numbers of the 10,000 possible.

If there is a pair, there are 12 ways to cover AABC (10*9*8), and a total of 3 possible pairs, AABC, ABBC, and ABCC. Though there are 720 families of AABC, the three possible pairs cause the number of families to be divided by 2 due to repeating families. There are 360 families of 12 for a total of 4,320 numbers.

If there are two pairs AABB (10*9) there are 45 families (remember that AABB and BBAA are the same family, so divide 90 by 2) * 6 ways for a total of 270 numbers.

If there is a triple AAAB there are 90 families * 4 ways (the placement of B) for a total of 360 numbers (here BBBA is not the same as AAAB).

All 4 numbers the same AAAA have 10 families of 1 way for a total of 10 numbers

The sum of all these families * ways is 10,000. The total number of tickets need to cover all 10,000 is (210+360+45+90+10) = 715 tickets

Well if Hot Blondes niece Is still waiting for the answer she will be in 9th grade by now.

Sorry dude, 6 graders don't know what a hypergeometric distribution is (which is actually what you are doing)! What you outline is no different than calculating poker hands, which isn't covered till an introductory probability and statistics course in college...

jet

Most 9th graders today still can't "solve" this problem. I bet she's still waiting.

As for a "silver" lining, if she develops into a "hot blonde" maybe she will get the answer from somebody else ;)

jet