Permutations and Combinations Questions

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.

There are 7 letters in the word DRAUGHT, the two vowels are A and U. Since, the vowels are not to be separated, AU can be considered as one entity. Therefore, the number of letters will be 6 instead of 7. The permutations will be P(6,6) = 6! ways.

But the two vowels A and U can be arranged in two ways, i.e. AU and UA. The required number of arrangements = 2!.6! = 1440 ways.

There are 4 books on fairy tales and they have to be put together. They can be arranged in 4! ways.

Similarly, there are 5 novels.They can be arranged in 5! ways.

And there are 3 plays.They can be arranged in 3! ways.

So, by the counting principle all of them together can be arranged in 4!´5!´3! ways = 17280

Choose 1 person for the single room & from the remaining choose 2 for the double room & from the remaining choose 4 people for the four person room, 

 Then, ⁷C₁ x ⁶C₂ x ⁴C₄ 

_{= 7 x 15 x 1 = 105}

4 novels can be selected out of 5 in 5C4 ways.

2 biographies can be selected out of 4 in 4C2 ways.

Number of ways of arranging novels and biographies = 5C4*4C2  = 30

After selecting any 6 books (4 novels and 2 biographies) in one of the 30 ways, they can be arranged on the shelf in 6! = 720 ways.

By the Counting Principle, the total number of arrangements = 30 x 720 = 21600

There are seven positions to be filled.

The first position can be filled using any of the 7 letters contained in PROBLEM.

The second position can be filled by the remaining 6 letters as the letters should not repeat.

The third position can be filled by the remaining 5 letters only and so on.

Therefore, the total number of ways of rearranging the 7 letter word = 7*6*5*4*3*2*1 = 7! 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

In a 3 digit number one’s place can be filled in 5 different ways with (0,2,4,6,8)

10’s place can be filled in 10 different ways

100’s place can be filled in 9 different ways

There fore total number of ways = 5X10X9 = 450