10. If integers $a, b, c$ satisfy: $0 \leqslant a \leqslant 10,0 \leqslant b \leqslant 10,0 \leqslant c \leqslant 10,10 \leqslant a+b+c \leqslant 20$, then the number of ordered triples $(a, b, c)$ that satisfy the condition is $\qquad$ groups.
When $c=0$, $10 \leqslant a+b \leqslant 20$, $(a, b)$ has $11+10+\cdots+1=66$ pairs; when $c=1$, $9 \leqslant a+b \leqslant 19$, $(a, b)$ has $10+11+10+\cdots+2=75$ pairs; when $c=2$, $8 \leqslant a+b \leqslant 18$, $(a, b)$ has $9+10+11+10+\cdots+3=82$ pairs; $\cdots$; when $c=10$, $0 \leqslant a+b \leqslant 10$, $(a, b)$ has $11+10+\cdots+1=66$ pairs. Therefore, the number of ordered triples $(a, b, c)$ that satisfy the condition is $2 \times(66+75+82+87+90)+91=891$.