Question 385820
10 hours can be sum of either of these set (according to question)...

( 5,5,0 ), (5,1,4), (5,2,3), (4,2,4) and (4,3,3)

now,

(5,5,0) can be arrange in 3 weeks by  3!/2! ways = 3

similarly, for(5,1,4)............ ....by 3! ways = 6 

......for (5,2,3) ........................ 3! ways = 6
     
......for (4,2,4).......................3!/2! ways = 3

 .....for (4,3,3)........................3!/2! ways = 3


total no. of ways.................................=  21  (sum of above)


so, total 21 ways......