Mistakes give rise to Problems- 12

Actually problem is to find solutions of $\displaystyle a^{b^{c}}= a^{bc}$

We break this in 3 cases, (i) $a=0$ , (ii) $a=1$ and (iii) $a\in [2,10]$

(i) $a=0$ , the case is true for all permitted $(b,c)$ , giving $100$ triples $(a,b,c)$

(ii) $a=1$ , the case is true for all permitted $(b,c)$ giving $100$ triples $(a,b,c)$

(iii) $a\in [2,10]$ , we have the case $b^c=bc$ and this is true if $c=1$ , giving $10$ pairs $(b,c)$ and $a$ has 9 choices, gives $90$ triples $(a,b,c)$

If $c\neq 1$ , only and only case is $b=c=2$ where we get $2^2=2\times 2=4$ again, $a$ has 9 choices, giving $9$ triples $(a,b,c)$

So as a whole there are $100+100+90+9= \boxed{299}$ triples following needed thing.