Let number of Rose plants be a.
Let number of marigold plants be b.
Let number of Sunflower plants be c.
According to question,
20a + 5b + 1c = 1000 - - - - - - (1)
a + b + c = 100 - - - - - - - - - - (2)
Solving above two equations by eliminating c, 19a + 4b = 900
b = $$\frac{{900 - 19a}}{4}$$   = $$225 - \frac{{19a}}{4}$$   - - - - - - - (3)
b being number of plants, is a positive integer, and is less than 99, as each of or two types have at least one plant in combination i.e .
0 < b < 99 - - - - - - - (4)
Substituting (3) in (4),
0 < 225 - $$\frac{{19a}}{4}$$ < 99
⇒ 225 < -$$\frac{{19a}}{4}$$ < (99 - 225)
⇒ 4 × 225 > 19a > 126 × 4
⇒ $$\frac{{900}}{{19}}$$ > a > 504
a is integer between 47 and 27 - - - - - - - - (5)
From (3), 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 or values of a, we get desired solution:
a=36,b=54,c=10
a=40,b=35,c=25
a=44,b=16,c=40
Three solutions are possible.

Since, all points are equally spaced; hence area of all convex pentagons will be same.

A team of 6 members has to be selected from 10 players. This can be done in ¹⁰C₆ or 210 ways.
Now, captain can be selected from se 6 players in 6 ways.
Therefore, total ways selection can be made is 210 × 6 = 1260

Alternatively
We can select 5 member team out of 10 in ¹⁰C₅ ways = 252 ways.
The captain can be selected from amongst remaining 5 players in 5 ways.
Therefore, total ways selection of 5 players and a captain can be made,
= 252 × 5
= 1260

No. of ways in which 10 paper can arranged is 10! Ways.
When best and worst papers come toger, regarding two as one paper, we have only 9 papers.
These 9 papers can be arranged in 9! Ways.
And two papers can be arranged mselves in 2! Ways.
No. of arrangement when best and worst paper do not come toger,
= 10! - 9! × 2!
= 9!(10 - 2)
= 8 × 9!

3 prize among 5 students can be distributed in,
⁵C₃ ways = 10 ways.

If two P’s were distinct (y could have different subscripts and colours), number of possible permutations would have been 5! = 120

For example let us consider one permutation: P₁LEAP₂

Now if we permute P’s amongst m we still get same word PLEAP. The two P’s can be permuted amongst m in 2! ways.

We were counting P₁LEAP₂ and P₂LEAP₁ as different arrangements only because we were artificially distinguishing between two P’s Hence number of different five letter words that can be formed is:
= $$\frac{{5!}}{{\left(1!\right) \left(2!\right) \left(1!\right) \left(1!\right)}}$$
= 60

The word 'OPTICAL' contains 7 different letters.
When vowels OIA are always toger, y can be supposed to form one letter.
Then, we have to arrange letters PTCL (OIA).
Now, 5 letters can be arranged in 5! = 120 ways.
The vowels (OIA) can be arranged among mselves in 3! = 6 ways.
Therefore Required number of ways = (120 x 6) = 720

The number of triangle can be formed by 10 points = ¹⁰C₃
Similarly, number of triangle can be formed by 4 points when no one is collinear = ⁴C₃
In question, given 4 points are collinear, Thus, required number of triangle can be formed,
= ¹⁰C₃ - ⁴C₃
= 120 - 4
= 116

Two tallest boys can be arranged in 2! ways, rest 18 can be arranged in 18! ways.
Girls can be arranged in 6! ways.
Total number of ways in which all students can be arranged,
= 2! × 18! × 6!
= 18! × 1440

Note:
N! = N × (N - 1) × (N - 2) × . . . . × 1
So,
18! = 18 × 17 × 16 × 15 . . . . . . . . × 1

The given word contains 8 letters of which E is taken 3 times.
∴ Required number of ways
$$\eqalign{ & = \frac{{8!}}{{3!}} \cr & = \frac{{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{6} \cr & = 6720 \cr} $$