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