Help please

A short code in C++ found:

276 odd triplets.

I will assume that they cannot be repeated.

there are (47-1)/2 = 23 odd positive numbers less than 47

43+4         1,3

41+6         1,5       

39+8          1,7         3,5

37+10        1,9         3,7        

35+12         1,11       3,9        5,7

33+14          1,13      3,11      5,9    

31+16          1,15      3,13      5,11      7,9

29+18           1,17     3,15      5,13     7,11       

27+20           1,19     3,17      5,15      7,13      9,11

25+22           1,21                                            9,13

------------------------------------------------------------------------------------

23+24          21, 3     19,5       17,7    15,9    13,11  

21+26          19,7      17,9       15,11    

19+28          17,11     15,13     

1+1+2+2+3+3+4+4+5+5     5+3+2  = 1+1+2+2+3+3+4+4+5+5+5+3+2 = 40

I get 40 odd triples with no repeats.    

Edit: I have changed this from 39 to 40 as there was an addition error at the very end.

Little numbers always get the best of me  

hmmm

let \(x=2a+1, y = 2b+1, z=2c+1\), so

\(2a+1+ 2b+1+2c+1=47\\ 2a+2b+2c=44\\ a+b+c=22\)

The problem reduces to:

"How many nonnegative ordered triples (a, b, c) satisfy \(a+b+c=22\)?"

which is a direct application of stars and bars, having an answer of \({22+3-1 \choose 3-1} = \frac{24!}{2!22!}=\boxed{276}\)

I'm getting the same answer as guest

Hi Textot,

Thanks for your contribution.

I really liked the way you started your answer.

You are of course correct,   

a+b+c = 22

where a,b, and c are ordered integers all greater or equal to 0

will have exactly the same number of solutions.

Your use of stars and bars was also interesting but unfortunately, it was not appropriate for this problem as it does not take into account that the triplet must be ordered.

This means that your answer is somewhere in the order of 3! or 6 times too big.

I used your simplification of  a+b+c=22

and made a probability contour map of it.

Instead of a,b, and c

I used

x,    y-x    and    22-y      (if you add these up you get 22)

Now these restraints come into play

\(x\ge0\qquad (1)\\~\\ y-x>x\\ y\ge2x\qquad(2)\\~\\ 22-y\ge y-x\\ -2y\ge -x-22\\ 2y\le x+22\\ y\le \frac{x}{2}+11\qquad (3)\\~\\ 22-y<22\\ y\ge0\qquad (4) \)

I graphed all these here      https://www.geogebra.org/classic/nzxqxyyv

this is the pic.

Every point in this region will give a triplet solution.

LaTex:

x\ge0\qquad (1)\\~\\
y-x>x\\
y\ge2x\qquad(2)\\~\\
22-y\ge y-x\\
-2y\ge -x-22\\
2y\le x+22\\
y\le \frac{x}{2}+11\qquad (3)\\~\\
22-y<22\\
y\ge0