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

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 find number of permutations of 4 objects out of 6 objects.

This number is 6P4= 360

Therefore, cards can be sent in 360 ways.

The bus fromA to B can be selected in 3 ways.

The bus from B to C can be selected in 4 ways.

The bus from C toD can be selected in 2 ways.

The bus fromD to E can be selected in 3 ways.

So, by the General Counting Principle, one can travel fromA to E in 3 x 4 x 2 x 3 ways = 72

Kernel follows Round Robin scheme choosing a swap device among the multiple swap devices in Unix System V.