The 5 letter word can be rearranged in 5!=120 Ways without any of the letters repeating.

The first 24 of these words will start with A.

Then the 25th word will start will CA _ _ _.
The remaining 3 letters can be rearranged in 3!=6 Ways. i.e. 6 words exist that start with CA.

The next word starts with CH and then A, i.e., CHA _ _.
The first of the words will be CHAMS. The next word will be CHASM.

Therefore, the rank of CHASM will be 24+6+2= 32

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.

As the toys are distinct and not identical,

For each of the 8 toys, we have three choices as to which child will receive the toy. Therefore, there are  5 8 ways to distribute the toys.

Hence, it is 5 8 and not 8 5.

Given word is THERAPY.

Number of letters in the given word = 7

Number of vowels in the given word = 2 = A & E

Required number of different ways, the letters of the word THERAPY arranged such that vowels always come together is

6! x 2! = 720 x 2 = 1440.

Here the order of choosing the elements doesn?t matter and this is a problem in combinations.

We have to find the number of ways of choosing 4 elements of this set which has 11 elements.

This can be done in 11 C 4 ways = 330 ways

There are 7 letters in the word Bengali of these 3 are vowels and 4 consonants.

There are 4 odd places and 3 even places. 3 vowels can occupy 4 odd places in 4 P 3 ways and 4 constants can be arranged in  4 P 4 ways.

Number of words = 4 P 3  x  4 P 4= 24 x 24 = 576

From 5 consonants, 3 consonants can be selected in  5 C 3 ways.

From 4 vowels, 2 vowels can be selected in  4 C 2 ways.

Now with every selection, number of ways of arranging 5 letters is 5 P 5ways.

Total number of words =  5 C 3 * 4 C 2 * 5 P 5

                                = 10x 6 x 5 x 4 x 3 x 2 x 1= 7200

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}

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  C 3 6 ways = 6x5x4 = 120 ways.

Number of words with 5 letters from given 9 alphabets formed =  9 5

Number of words with 5 letters from given 9 alphabets formed such that no letter is repeated is =  9 P 5

Number of words can be formed which have at least one letter repeated =   9 5 - 9 P 5

= 59049 - 15120

= 43929