The correct option is A 140
The objects will either be in a distribution of 1,1,1,1,3 or 1,1,1,2,2.
Note that it does not matter what order the objects are in, only which objects are grouped together. For the first case, there are 7C3 ways to choose the grouping of 3 objects. Then, in the second case, there are 7C4 ways to select 4 objects out of the 7, and then there are 4C22 = 3 ways to divide these 4 objects into 2 groups of 2. The total number of ways to arrange the objects is 7C3+3∗7C4=140