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

RAINBOW

In given word OLIVER there are 3 vowels E, I & O. These can be arranged in only one way as dictionary order E, I & O.

There are 6 letters in thegiven word.

First arrange 3 vowels.

This can be done in 6C3 ways and that too in only one way.(dictionary order E, I & O)

Remaining 3 letters can be placed in 3 places = 3! ways

Total number of possible ways of arranging letters of OLIVER = 3! x C36 ways = 6x5x4 = 120 ways.

We are having with digits 2, 3, 4, 5 & 6 and numbers greater than 4000 are to be formed, no digit is repeated.

The number can be 4 digited but greater than 4000 or 5 digited.

Number of 4 digited numbers greater than 4000 are
4 or 5 or 6 can occupy thousand place => 3 x P34 = 3 x 24 = 72. 

5 digited numbers = P55 = 5! = 120 

So the total numbers greater than 4000 = 72 + 120 = 192

NUMERICAL has 9 positions in which 2, 4, 6, 8 are even positions.

And it contains 5 consonents i.e, N, M, R, C & L. Hence this cannot be done as 5 letters cannot be placed in 4 positions.

Therefore, Can't be determined.

The following arrangements satisfy all 3 conditions.

Arrangement 1: 3 books in a row; 12 rows.

Arrangement 2: 4 books in a row; 9 rows.
Arrangement 3: 6 books in a row; 6 rows.
Arrangement 4: 9 books in a row; 4 rows.
Arrangement 5: 12 books in a row; 3 rows.

Therefore, the possible arrangements are 5.

Given 11 questions of type True or False

Then, Each of these questions can be answered in 2 ways (True or false)

Therefore, no. of ways of answering 11 questions = 211 = 2048 ways.

we can select the 5 member team out of the 8 in 8C5 ways = 56 ways.

The captain can be selected from amongst the remaining 3 players in 3 ways.

Therefore, total ways the selection of 5 players and a captain can be made = 56x3 = 168 ways.

(or)

Alternatively, A team of 6 members has to be selected from the 8 players. This can be done in 8C6 or 28 ways.

Now, the captain can be selected from these 6 players in 6 ways.

Therefore, total ways the selection can be made is 28x6 = 168 ways.

The given digits are 1, 2, 3, 5, 7, 9  

Number of even numbers = ⁵P₃ = 60.

Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.

Probability that both marbles are blue = ³C₂/¹⁵C₂ = (3 x 2)/(15 x 14) = 1/35

Probability that both are yellow = ²C₂/¹⁵C₂ = (2 x 1)/(15 x 14) = 1/105

Probability that one blue and other is yellow = (³C₁ x ²C₁)/¹⁵C₂ = (2 x 3 x 2)/(15 x 14) = 2/35

Required probability = 1/35 + 1/105 + 2/35 = 3/35 + 1/105 = 1/35(3 + 1/3) = 10/(3 x 35) = 2/21