In the below word how many words are there in which R and W are at the end positions?

RAINBOW

Sum of 4 digit numbers = (2+4+6+8) x P33 x (1111) = 20 x 6 x 1111 = 133320 

Sum of 3 digit numbers = (2+4+6+8) x P23 x (111) = 20 x 6 x 111 = 13320 

Sum of 2 digit numbers = (2+4+6+8) x P13 x (11) = 20 x 3 x 11 = 660 

Sum of 1 digit numbers = (2+4+6+8) x P03 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.

C19+C18+C16 = 23 will be answer.

Given G + B= 41 and B = G-1
Hence, G = 21 and B = 20
Now we have 2 options,
1G and 3M
(or)
3G and 1M
(2G and 2M or 0G and 4M or 4G and oM are not allowed),

Total : C121×C320+C321×C120= 50540 ways.

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

Given there are 9 flavours of ice creams.

Each child takes the combination of two flavours and no two children will have the same combination

This can be done by  ways i.e children.

Number of children=C29  = 9 x 8 / 1x 2 = 36.

No. of letters in the word = 6
No. of 'E' repeated = 2
Total No. of arrangement = 6!/2! = 360

Number of mens = 8

Number of womens = 5

Different ways could couples be picked = C16× 9C1 = 9 x 6 = 54 ways.

Let last digit is 2
when second last digit is 4 remaining 4 digits can be filled in 120 ways, similarly second last digit is 6 remained 4 digits can be filled in 120 ways.
so for last digit = 2, total numbers=240

Similarly for 4 and 6
When last digit = 4, total no. of ways =240
and last digit = 6, total no. of ways =240
so total of 720 even numbers are possible.