Bars and stars approach can also be used, if we restrict one star each in 3 partitions we will have 11C9 combinations i.e 55 . From this we should subtract case when any partition has greater than 7 stars. 7 star in x1 , x2 1 star and x3 1 star, select remaining 3 stars from 3 star+2 bars i.e 5 C 3 .This will be 3 times as we can place 7 stars in x1 or x2 or x3. So 11C9 - 3(5C3) = 25. But I think generating functions process is much easier.