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
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.
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
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
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
=
The total number of ways of forming the group of ten representatives is ²²C??.
The total number of ways of forming the group that consists of no seniors is ¹?C?? = 1 way
The required number of ways = ²²C?? - 1
n items of which p are alike of one kind, q alike of the other, r alike of another kind and the remaining are distinct can be arranged in a row in n!/p!q!r! ways.
The letter pattern 'MESMERISE' consists of 10 letters of which there are 2M's, 3E's, 2S's and 1I and 1R.
Number of arrangements =
Two men, three women and one child can be selected in ?C? x ?C? x ?C? ways = 600 ways.
Given word is GLACIOUS has 8 letters.
=> C is fixed in one of the 8 places
Then, the remaining 7 letters can be arranged in 7! ways = 5040.
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