There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.

Let us mark these positions as under: 

                                                      (1) (2) (3) (4) (5) (6) 

Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.  

Number of ways of arranging the vowels = 3 P 3 = 3! = 6.

Also, the 3 consonants can be arranged at the remaining 3 positions. 

Number of ways of these arrangements =  3 P 3 = 3! = 6. 

Total number of ways = (6 x 6) = 36.

The first letter from the right can be chosen in 26 ways because there are 26 alphabets.

Having chosen this, the second letter can be chosen in 26 ways

The first two letters can chosen in 26 x 26 = 676 ways

Having chosen the first two letters, the third letter can be chosen in 26 ways.

All the three letters can be chosen in 676 x 26 =17576 ways.

It implies that the maximum possible number of five letter palindromes is 17576 because the fourth letter is the same as the second letter and the fifth letter is the same as the first letter.

Each of the 10 letters can be posted in any of the 5 boxes.

So, the first letter has 5 options, so does the second letter and so on and so forth for all of the 10 letters.

i.e. 5*5*5*?.*5 (upto 10 times) = 5 ^ 10.

Given letters are k, l, m, n, o = 5

number of letters to be in the words = 3

Total number of words that can be formed from these 5 letters taken 3 at a time without repetation of letters =  5 P 3 ways .

? 5 P 3 = 5 x 4 x 3 = 60 words .

15! - 14! - 13!

= (15 × 14 × 13!) - (14 × 13!) - (13!)

= 13! (15 × 14 - 14 - 1)

= 13! (15 × 14 - 15)

= 13! x 15 (14 - 1)

= 15 × 13 × 13!

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

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

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

Probability that one blue and other is yellow = (³C? x ²C?)/¹?C? = (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

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 = 2 11 = 2048 ways.

The number of letters in the given word CREATIVITY = 10

Here T & I letters are repeated

=> Number of Words that can be formed from CREATIVITY = 10!/2!x2! = 3628800/4 = 907200