Four tenths = 0.4

Five thousandths = 0.005

The average is (0.4 + 0.005)/2 = 0.2025

The weight will be 250g plus (1.55 - 0.65)/0.10 units of 100g

250 + 900 = 1150

This is the maximum weight that can be sent at that price. But, weights exceeding

250 + 800 will also get charged this amount (that is what the ‘part thereof’ implies).

Hence a package weighing 1145 will be charged $1145

The possible outcomes that satisfy the condition of "at least one house gets the wrong package" are:
One house gets the wrong package or two houses get the wrong package or three houses get the wrong package.

We can calculate each of these cases and then add them together, or approach this problem from a different angle.
The only case which is left out of the condition is the case where no wrong packages are delivered.

If we determine the total number of ways the three packages can be delivered and then subtract the one case from it, the remainder will be the three cases above.

There is only one way for no wrong packages delivered to occur. This is the same as everyone gets the right package.

The first person must get the correct package and the second person must get the correct package and the third person must get the correct package.
 1×1×1=1

Determine the total number of ways the three packages can be delivered.
 3×2×1=6

The number of ways at least one house gets the wrong package is:
  6−1=5
Therefore there are 5 ways for at least one house to get the wrong package.

Let the number of Rose plants be ‘a’.
Let number of marigold plants be ‘b’.
Let the number of Sunflower plants be ‘c’.
20a+5b+1c=1000; a+b+c=100

Solving the above two equations by eliminating c,
19a+4b=900

b = (900-19a)/4 

b = 225 - 19a/4----------(1)

b being the number of plants, is a positive integer, and is less than 99, as each of the other two types have at least one plant in the combination i.e .:0 < b < 99--------(2)

Substituting (1) in (2),

 0 < 225 - 19a/4 < 99

225 <  -19a/4 < (99 -225)

=> 4 x 225 > 19a > 126 x 4

=> 900/19 > a > 505

a is the integer between 47 and 27 ----------(3)
From (1), it is clear, a should be multiple of 4.

Hence possible values of a are (28,32,36,40,44)

For a=28 and 32, a+b>100
For all other values of a, we get the desired solution:
a=36,b=54,c=10
a=40,b=35,c=25
a=44,b=16,c=40

Three solutions are possible.