There are 3 bags, in 1st there are 9 Mangoes, in 2nd 8 apples & in 3rd 6 bananas. There are how many ways you can buy one fruit if all the mangoes are identical, all the apples are identical, & also all the Bananas are identical ?

Correct Answer: 32640

Explanation:

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

Correct Answer: None of these

Explanation:

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

Correct Answer: 3

Explanation:

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.

Correct Answer: 4 x 3^4

Explanation:

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

Correct Answer: 252

Explanation:

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

Correct Answer: 147320

Explanation:

Sum of 4 digit numbers = (2+4+6+8) x  P 3 3 x (1111) = 20 x 6 x 1111 = 133320 

Sum of 3 digit numbers = (2+4+6+8) x  P 2 3 x (111) = 20 x 6 x 111 = 13320 

Sum of 2 digit numbers = (2+4+6+8) x  P 1 3 x (11) = 20 x 3 x 11 = 660 

Sum of 1 digit numbers = (2+4+6+8) x  P 0 3 x (1) = 20 x 1 x 1 = 20 

 

Adding All , Sum = 147320

Correct Answer: C

Explanation:

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
= 2 4 - 1 ........(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
= 2 3 - 1........(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
= 2 5..............(C)

 

From (A), (B) and (C), required number of ways
=   2 5 2 4 - 1 2 3 - 1

Correct Answer: ²²C?? - 1

Explanation:

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

Correct Answer: 9!/(2!)^{2}x3!

Explanation:

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 =  9 ! ( 2 ! ) 2 × 3 !

Correct Answer: 600

Explanation:

Two men, three women and one child can be selected in ?C? x ?C? x ?C? ways = 600 ways.