Here Only $4$ cases are possible .
$ Case-1 $ When All Three numbers are $-ve$
in this case only $4$ combinations possible and we can choose any number and if multiply those it will $+ve$ number .
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$ Case-2 $ When $1$ $+ve$ , $2$ $-ve$
in this case when $0$ willl be choosen in the triplets After multiplying it will give $0$ which will anyway greater then the other $-ve$ number. So in this with $0 \begin{Bmatrix} -4,-3 & \\ -4,-1& \\ -4,-2& \\ -3,-2& \\ -3,-1 & \\ -2,-1& \end{Bmatrix}$ these possible combination will give right Tuples. and with Remaining numbers 1,2.. and so on We can't take Coz it will give One -ve number which is voilating the condition. So 6 from here .
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$ Case-3 $ When $1$ $+ve$ , $2$ $-ve$
These all also will voilating the condiition . So can't choose.
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$ Case-4 $ When All $+ve$
only ${2,3,4}$ trplet is not voilation the condition So only 1 from this case.
Total $11$ possible combinations only.
