6. Given the sets $A=\{1,2,3,4,5,6\}, B=$ $\{6,7,8,9\}$, by selecting 3 elements from $A$ and 2 elements from $B$, we can form $\qquad$ new sets with 5 elements.
6. Selecting 3 elements from set $A$ has $C_{6}^{3}$ ways, and selecting 2 elements from set $B$ without 6 has $C_{3}^{2}$ ways, resulting in a total of $C_{6}^{3} \cdot C_{3}^{2}$ ways; selecting 3 elements from set $A$ without 6 has $C_{5}^{3}$ ways, and selecting 1 element from set $B$ with 6 has $C_{3}^{1}$ ways, resulting in a total of $C_{5}^{3} \cdot C_{3}^{1}$ ways. Therefore, the number of sets that meet the conditions is $C_{6}^{3} \cdot C_{3}^{2} + C_{5}^{3} \cdot C_{3}^{1} = 90$ (sets).