The number of ways in which the letters A, B, C, D, E can be placed in the 8 boxes of the figure (arranged in rows of size 3, 3, 2) so that no row remains empty and at most one letter can be placed in a box, is:
Solutions · 1
Official solution
Let x,y,z be the number of letters placed in rows of size 3,3,2 respectively, with x+y+z=5, 1≤x≤3,1≤y≤3,1≤z≤2.
Valid (x,y,z) and box-choice counts (x3)(y3)(z2):
- (3,1,1): (33)(13)(12)=6
- (1,3,1): (13)(33)(12)=6
- (2,2,1): (23)(23)(12)=18
- (2,1,2): (23)(13)(22)=9
- (1,2,2): (13)(23)(22)=9
Total ways to choose boxes =6+6+18+9+9=48
Then arrange the 5 distinct letters into the 5 chosen boxes: 5!=120.
\[\text{Total} = 48\times120 = 5760\]