n(E) = 5C3 + 4C3 + 3C3 = 10 + 4 + 1 = 15n(S) = 12C3 = 220

They have to be arranged in the following way :

                                                                    L  T  L  T  L  T  L  T  L

The 5 lions should be arranged in the 5 places marked ?L?.

This can be done in 5! ways.

The 4 tigers should be in the 4 places marked ?T?.

This can be done in 4! ways.

Therefore, the lions and the tigers can be arranged in 5!´ 4! ways = 2880 ways.

We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only). 

Required number of ways=  7 C 3 * 6 C 2 + 7 C 4 * 6 C 1 + 7 C 5 = 756.

We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).

Required number of ways= 3 C 1 * 6 C 2 + 3 C 2 * 6 C 1 + 3 C 3 = (45 + 18 + 1) =64

The solution to this problem involves calculating a factorial. Since we want to know how 13 cards can be arranged, we need to compute the value for 13 factorial.

13! = (1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13) = 6,227,020,800

Since there are 5 cups of each kind,prepared with milk or tea leaves added first,are identical hence,total number of different people ways of presenting the cups to the expert is 10!/(5! x 5!)= 252

The remainder on the first card can be 0,1,2 or 3 i.e 4 possibilities.
The remainder of the number on the next card when divided by 4 can have 3 possible values (except the one occurred earlier).

For each value on the card the remainder can have 3 possible values.

The total number of possible sequences is: 4*3^4

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.

100 C 10 + 90 C 20 + 70 C 30 + 40 C 40 = 100 ! 10 ! x 20 ! x 30 ! x 40 !

We know that zero can't be in hundreds place. But let's assume that our number could start with zero.

The formula to find sum of all numbers in a permutation is

111 x no of ways numbers can be formed for a number at given position x sum of all given digits

No of 1 s depends on number of digits

So,the answer us

111 x 20 x (0+1+2+3+4+5) = 33300

We got 20 as follows. If we have 0 in units place we can form a number in 4*5 ways. This is for all numbers. So we have substituted 20 in formula.

Now, this is not the final answer because we have included 0 in hundreds place. so we have to remove the sum of all numbers that starts with 0.

This is nothing but the sum of all 2 digits numbers formed by 1 2 3 4 5. Because 0 at first place makes it a 2 digit number.

So the sum for this is 11 x 4 x (1+2+3+4+5).
=660

Hope u understood why we use 4. Each number can be formed in 4x1 ways

So, the final answer is 33300-660 = 32640