How many different ways can Hector put 5 tulips in 2 vases?
Ordered pair (V1,V2) represents the number of tulips in the two vases. V1 and V2 identify the vases.
The set of possibilities are: (5,0), (4,1). (3,2), (2,3), (1,4), (0,5) if the tulips are identical.
If the tulips are distinct: A, B, C, D, E, the permutations are:
(5,0): (ABCDE,0) 
(4.1): (BCDE,A), (ACDE,B), (ABDE,C), (ABCE,D), (ABCD,E) 
(3,2): (ABC,DE), (ABD,CE), (ABE,CD), (ACD,BE), (ACE,BD), (ADE,BC), (BCD,AE), (BCE,AD), (BDE,AC), (CDE,AB) 
(2,3): (DE,ABC), (CE,ABD), (CD,ABE), (BE,ACD), (BD,ACE), (BC,ADE), (AE,BCD), (AD,BCE), (AC,BDE), (AB,CDE) 
(1,4): (A,BCDE), (B,ACDE), (C,ABDE), (D,ABCE), (E,ABCD) 
(0,5): (0,ABCDE)  [32 ways in total=1+5+10+10+5+1]
If the tulips in each vase can be placed in different orders, then there are more permutations: 2 ways of arranging 2 tulips; 6 ways of arranging 3; 24 ways for 4; 120 ways for 5. For example, for (BC,ADE) there are 2 ways to arrange BC, and 6 ways to arrange ADE. So (BC,ADE) spawn 2*6=12 arrangements. Since there are 10 (2,3) arrangements, each of these spawns 12, giving a total of 120 possible ways. For (1,4) each spawns 24, making 5*24=120 possible ways. For (0,5) again we have 120 ways. Since there are 6 rows above the grand total would be 6*120=720 ways.
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