Since two of the eight first class petty officers are to fill two different offices, we write 8P2=56

Then, two of the six second class petty officers are to fill two different offices; thus, we write 6P2 =30

The principle of choice holds in this case; therefore, the members have 56 x 30 = 1680 ways to select the required office holders

You need two points to draw a line. The order is not important. Line AB is the same as line BA. The problem is to select 2 points out of 3 to draw different lines. If we proceed as we did with permutations, we get the following pairs of points to draw lines.

AB , AC

BA , BC

CA , CB

There is a problem: line AB is the same as line BA, same for lines AC and CA and BC and CB.

The lines are: AB, BC and AC ; 3 lines only.

So in fact we can draw 3 lines and not 6 and that's because in this problem the order of the points A, B and C is not important.

High level data models provide the concepts that are close to the way many users perceive data

ABACUS is a 6 letter word with 3 of the letters being vowels.

If the 3 vowels have to appear together, then there will 3 other consonants and a set of 3 vowels together.

These 4 elements can be rearranged in 4! Ways.

The 3 vowels can rearrange amongst themselves in 3!/2! ways as "a" appears twice.

Hence, the total number of rearrangements in which the vowels appear together are (4! x 3!)/2!

Database provides necessary means to achieve this abstraction. By structure of a database we mean the data types, relations, and constraints that should hold on the data.

The first letter is E and the last one is R.

Therefore, one has to find two more letters from the remaining 11 letters.

Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters.

The second and third positions can either have two different letters or have both the letters to be the same.

Case 1: When the two letters are different. One has to choose two different letters from the 8 available different choices. This can be done in 8 * 7 = 56 ways.

Case 2: When the two letters are same. There are 3 options - the three can be either Ns or Es or As. Therefore, 3 ways.

Total number of possibilities = 56 + 3 = 59

Out of 26 alphabets two distinct letters can be chosen in 26P2 ways. Coming to numbers part, there are 10 ways.(any number from 0 to 9 can be chosen) to choose the first digit and similarly another 10ways to choose the second digit. Hence there are totally 10X10 = 100 ways. 

Combined with letters there are 6P2 X 100 ways = 65000 ways to choose vehicle numbers.

1 million distinct 3 digit initials are needed.

Let the number of required alphabets in the language be ‘n’.

Therefore, using ‘n’ alphabets we can form n * n * n = n3 distinct 3 digit initials.

Note distinct initials is different from initials where the digits are different.

For instance, AAA and BBB are acceptable combinations in the case of distinct initials while they are not permitted when the digits of the initials need to be different.

This n3 different initials = 1 million 

i.e. n3=106  (1 million = 106)

  => n = 102 = 100

Hence, the language needs to have a minimum of 100 alphabets to achieve the objective.