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How many ordered quadruples (a,b,c,d) satisfy a+b+c+d=18, where a,b,c,d are odd positive integers?

How many ordered quadruples (a,b,c,d) satisfy a+b+c+d=18, where a,b,c,d are integers such that |a|,|b|,|c|,|d| are each at most 10?

For some reason, I am having tons of difficulty with these problems!! How do I start? Can you explain it?

Mathy Person
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1 Answers1

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For the first, let our numbers be $2s+1,2t+1,2u+1, 2v+1$, where $s,t,u,v$ are all non-negative. These add up to $18$ if and only if $s+t+u+v=\frac{18-4}{2}=7$.

You know how to find the number of quadruples of non-negative integers with sum $7$.

André Nicolas
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  • Which question were you replying to? and 6C3, so 20? – Mathy Person Nov 01 '14 at 02:57
  • I tried out 20 and it was incorrect. Would it have been 6C3? – Mathy Person Nov 02 '14 at 16:54
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    I was replying to the first question. It is quadruples of non-negative integers. Usual stars and bars says $\binom{7+4-1}{4-1}$. – André Nicolas Nov 02 '14 at 17:13
  • Oh I see, I forgot about the "non-negative" part. Do you have a hint for the second question as well? – Mathy Person Nov 08 '14 at 19:35
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    Convert to $0$ to $20$ sum $58$ to make intuition clearer, then use Stars and Bars with Inclusion/Exclusion. A bit messy. We can alternately use generating functions, but the final coefficient calculation, unless done by computer, does not make this approach simpler. – André Nicolas Nov 08 '14 at 19:55