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

RAINBOW

Possibilities     Bowlers      Batsmen         Number of ways

                         6               9

         1              4                7              (6C4*9C7)

         2              5                6              6C5*9C6

         3              6                5              6C6*9C5

6C4*9C7 = 15 x 36 = 540

6C5*9C6 = 6 x 84 = 504

 6C6*9C5= 1 x 126 = 126

Total = 1170

A triangle is formed by joining any three non-collinear points in pairs.

There are 7 non-collinear points

The number of triangles formed = 7C3 = 35

Two horses A and B, in a race of 6 horses... A has to finish before B

if A finishes 1... B could be in any of other 5 positions in 5 ways and other horses finish in 4! Ways, so total ways 5*4!

if A finishes 2... B could be in any of the last 4 positions in 4 ways. But the other positions could be filled in 4! ways, so the total ways 4*4!

if A finishes 3rd... B could be in any of last 3 positions in 3 ways, but the other positions could be filled in 4! ways, so total ways 3*4!

if A finishes 4th... B could be in any of last 2 positions in 2 ways, but the other positions could be filled in 4! ways, so total ways... 2 * 4! 

if A finishes 5th .. B has to be 6th and the top 4 positions could be filled in 4! ways..

A cannot finish 6th, since he has to be ahead of B

Therefore total number of ways = 5*4! + 4*4! + 3*4! + 2*4! + 4! = 120 + 96 + 72 + 48 + 24 = 360

Selecting zero'A's= 1

Selecting one 'A's = 1

Selecting two 'A's = 1

Selecting three 'A's = 1

Selecting four 'A's = 1

Selecting five 'A's = 1

=> Required number ofways =6

Total number of alphabets = 10

so ways to arrange them = 10! 

Then there will be duplicates because 1st S is no different than 2nd S.

we have 4 Is 3 S and 2 Ps 

Hence number of arrangements = 10!/4! x 3! x 2! = 12600

Here clock-wise and anti-clockwise arrangements are same.

Hence total number of circular–permutations: 18P122*12 = 18!6*24

The toys are different; The boxes are identical 

If none of the boxes is to remain empty, then we can pack the toys in one of the following ways 

a. 2, 2, 1 

b. 3, 1, 1 

Case a. Number of ways of achieving the first option 2 - 2 - 1 

Two toys out of the 5 can be selected in 5C2 ways. Another 2 out of the remaining 3 can be selected in 3C2 ways and the last toy can be selected in 1C1 way. 

However, as the boxes are identical, the two different ways of selecting which box holds the first two toys and which one holds the second set of two toys will look the same. Hence, we need to divide the result by 2 

Therefore, total number of ways of achieving the 2 - 2 - 1 option is ways 5C2*3C2= 15 ways

Case b. Number of ways of achieving the second option 3 - 1 - 1

Three toys out of the 5 can be selected in 5C3 ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way.

Therefore, total number of ways of getting the 3 - 1 - 1 option is 5C3 = 10 = 10 ways.

Total ways in which the 5 toys can be packed in 3 identical boxes

= number of ways of achieving Case a + number of ways of achieving Case b= 15 + 10 = 25 ways.

The smallest number in the series is 1000, a 4-digit number.

The largest number in the series is 4000, the only 4-digit number to start with 4. 

The left most digit (thousands place) of each of the 4 digit numbers other than 4000 can take one of the 3 values 1 or 2 or 3.

The next 3 digits (hundreds, tens and units place) can take any of the 5 values 0 or 1 or 2 or 3 or 4.

Hence, there are 3 x 5 x 5 x 5 or 375 numbers from 1000 to 3999.

Including 4000, there will be 376 such numbers.