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.
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
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 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
We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).
Required number of ways= = (45 + 18 + 1) =64
We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
Required number of ways= = 756.
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
As in this problem , buying any fruit is different case , as buying apple is independent from buying banana. so ADDITION rule will be used.
= 23 will be answer.
Sum of 4 digit numbers = (2+4+6+8) x x (1111) = 20 x 6 x 1111 = 133320
Sum of 3 digit numbers = (2+4+6+8) x
x (111) = 20 x 6 x 111 = 13320
Sum of 2 digit numbers = (2+4+6+8) x
x (11) = 20 x 3 x 11 = 660
Sum of 1 digit numbers = (2+4+6+8) x
x (1) = 20 x 1 x 1 = 20
Adding All , Sum = 147320
It is explicitly given that all the 4 black balls are different, all the 3 red balls are different and all the 5 blue balls are different. Hence this is a case where all are distinct objects.
Initially let's find out the number of ways in which we can select the black balls. Note that at least 1 black ball must be included in each selection.
Hence, we can select 1 black ball from 4 black balls
or 2 black balls from 4 black balls.
or 3 black balls from 4 black balls.
or 4 black balls from 4 black balls.
Hence, number of ways in which we can select the black balls
= 4C1 + 4C2 + 4C3 + 4C4
=
........(A)
Now let's find out the number of ways in which we can select the red balls. Note that at least 1 red ball must be included in each selection.
Hence, we can select 1 red ball from 3 red balls
or 2 red balls from 3 red balls
or 3 red balls from 3 red balls
Hence, number of ways in which we can select the red balls
= 3C1 + 3C2 + 3C3
=
........(B)
Hence, we can select 0 blue ball from 5 blue balls (i.e, do not select any blue ball. In this case, only black and red balls will be there)
or 1 blue ball from 5 blue balls
or 2 blue balls from 5 blue balls
or 3 blue balls from 5 blue balls
or 4 blue balls from 5 blue balls
or 5 blue balls from 5 blue balls.
Hence, number of ways in which we can select the blue balls
= 5C0 + 5C1 + 5C2 + ? + 5C5
=
..............(C)
From (A), (B) and (C), required number of ways
=
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