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

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 11C4 ways = 330 ways

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 =
4%, m=
12, FV =
$10,000, Term=
3.5 years
Then n
=m  *Term
12(3.5)
42

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

Let 4 girls be one unit and now there are 6 units in all.

They can be arranged in 6! ways.

In each of these arrangements 4 girls can be arranged in 4! ways. 

Total number of arrangements in which girls are always together = 6! x 4!= 720 x 24 = 17280