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

i=j/m

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

The single equivalent payment will be PV + FV.
FV
= Future value of $10,000, 12 months later

$10,000 *(1.0075)/12

$10,938.07
PV=
Present value of $10,000, 24 months earlier

$10,000/(1.0075)24

$8358.31
The equivalent single payment is
$10,938.07 + $8358.31
= $19,296.38

i=j/m

PV=
FV(1+  i)^-n

We have to find number of permutations of 4 objects out of 6 objects.

This number is 6P4= 360

Therefore, cards can be sent in 360 ways.

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 word 'OPTICAL' contains 7 different letters.

When the vowels OIA are always together, they can be supposed to form one letter.

Then, we have to arrange the letters PTCL (OIA).

Now, 5 letters can be arranged in 5! = 120 ways.

The vowels (OIA) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.