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 4P3 ways and 4 constants can be arranged in 4P4 ways.

Number of words =4P3  x 4P4= 24 x 24 = 576

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

Considering vowels a, e, i as one letter, we can arrange 4+1 letters in 5! ways in each of which vowels are together. These 3 vowels can be arranged among themselves in 3! ways.

Total number of words = 5! x 3!= 120 x 6 = 720

Given:j=7% compounded semiannually making m=2 and i = j/m= 7%/2 = 3.5%
Let x represent the third payment. Then the second payment must be 2x.
PV1,PV2, andPV3 represent the present values of the first, second, and third payments.

Since the sum of the present values of all payments equals the original loan, then
PV1 + PV2  +PV3
=$4000 -------(1)

PV1
  =FV/(1 + i)^n
=$1000/(1.035)^4=
$871.44

At first, we may be stumped as to how to proceed for
PV2 and PV3. Let’s think about the third payment of x dollars. We can compute the present value of just $1 from the x dollars

pv=1/(1.035)^10=0.7089188

PV2
  =2x * 0.7089188 =
1.6270013x
PV3
  =x * 0.7089188=0.7089188x
Now substitute these values into equation ➀ and solve for x.
$871.442 + 1.6270013x + 0.7089188x
=$4000

2.3359201x
=$3128.558

x=$1339.326
Kramer’s second payment will be 2($1339.326)
=$2678.65, and the third payment will be $1339.33

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

Let the beds be numbered 1 to 7.

Case 1 : Suppose Anju is allotted bed number 1. 

Then, Parvin cannot be allotted bed number 2. 

So Parvin can be allotted a bed in 5 ways. 

After alloting a bed to Parvin, the remaining 5 students can be allotted beds in 5! ways.

So, in this case the beds can be allotted in 5´5!ways = 600 ways.

Case 2 : Anju is allotted bed number 7. 

Then, Parvin cannot be allotted bed number 6 

As in Case 1, the beds can be allotted in 600 ways.

Case 3 : Anju is allotted one of the beds numbered 2,3,4,5 or 6. 

Parvin cannot be allotted the beds on the right hand side and left hand side of Anju’s bed. For example, if Anju is allotted bed number 2, beds numbered 1 or 3 cannot be allotted to Parvin.

Therefore, Parvin can be allotted a bed in 4 ways in all these cases.

After allotting a bed to Parvin, the other 5 can be allotted a bed in 5! ways.

Therefore, in each of these cases, the beds can be allotted in 4´ 5! = 480 ways. 

The beds can be allotted in (2x 600 + 5 x 480)ways = (1200 + 2400)ways = 3600 ways

We have to arrange 6 books.

The number of permutations of n objects is n! = n. (n – 1) . (n – 2) ... 2.1

Here n = 6 and therefore, number of permutations is 6.5.4.3.2.1 = 720

Three digit number will have unit’s, ten’s and hundred’s place.

Out of 5 given digits any one can take the unit’s place.

This can be done in 5 ways. ...              (i)

After filling the unit’s place, any of the four remaining digits can take the ten’s place.

This can be done in 4 ways. ...              (ii)

After filling in ten’s place, hundred’s place can be filled from any of the three remaining digits.

This can be done in 3 ways. ... (iii) 

So,by counting principle, the number of 3 digit numbers = 5x 4 x 3 = 60

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.