Math puzzle: combinations of digits

AI Thread Summary
The discussion revolves around calculating the number of unique combinations of the digits 1 through 5, considering single to five-digit combinations without repetitions or reversals. The initial calculations suggested a total of 78 combinations, but this was incorrect. The correct total, as revealed later in the discussion, is 31 combinations. Participants emphasized the importance of understanding the methodology behind the calculations rather than just obtaining the final answer. The conversation highlights the value of mathematical reasoning and pattern recognition in solving such puzzles.
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My 10 year old daughter was given this maths puzzle and I'm sure to you guys it would be pretty easy.

You have 5 numbers 1,2,3,4,5

Ho many possible combinations are there possibe for 1 digit combos,2 digit combos, 3 digit combos, 4 digit combos, and 5 digit combos. The numbers cannot repeat at any time and you cannot have the reverse of the numbers also (e.g 345 and 543 as same 3 numbers used in this 3 digit combo.)

For single digit combos there is onviously only 5 combos 1-5 (5)
For 5 digit combos there is only one combination 1-5 (1) as any other combo still uses the same 5 numbers
I thought using 5*4*3*2*1 / 4+3+2+1 = (12) would be the correct answer for the 4 digit combos
and that 5*4*3*2*1 / 3 +2 +1 = (20) would be correct for 3 digit combos and that 5*4*3*2*1 / 2+1 = (40) for 2 digit combos.

So total combos 5 + 40 + 20 + 12 + 1 = 78 possible combos would be correct but it's not.

What is the correct answer and can you explain how it is done.

Thanks in advance
 
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She should start by listing all possibilities for 2 digits (including reversals), then excluding reversals, and try to deduce a pattern from that.
 
Logger said:
My 10 year old daughter was given this maths puzzle and I'm sure to you guys it would be pretty easy.

You have 5 numbers 1,2,3,4,5

Ho many possible combinations are there possibe for 1 digit combos,2 digit combos, 3 digit combos, 4 digit combos, and 5 digit combos. The numbers cannot repeat at any time and you cannot have the reverse of the numbers also (e.g 345 and 543 as same 3 numbers used in this 3 digit combo.)

For single digit combos there is onviously only 5 combos 1-5 (5)
For 5 digit combos there is only one combination 1-5 (1) as any other combo still uses the same 5 numbers
I thought using 5*4*3*2*1 / 4+3+2+1 = (12) would be the correct answer for the 4 digit combos
and that 5*4*3*2*1 / 3 +2 +1 = (20) would be correct for 3 digit combos and that 5*4*3*2*1 / 2+1 = (40) for 2 digit combos.

So total combos 5 + 40 + 20 + 12 + 1 = 78 possible combos would be correct but it's not.

What is the correct answer and can you explain how it is done.

Thanks in advance

For 4-digit combos, every combo is defined by the number missing. So, there are not 12 of those.
 
Logger said:
Somebody sent me the answer...31
You really think your daughter should only care about the numerical answer?
 
The link that I included on my previous post showed the formula and how it works and the proof. Do you really think a maths teacher would allow a child just come up with a numerical answer anyway? Thanks for taking the time to reply.Take care.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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