The solution to this problem involves calculating a factorial. Since we want to know how 13 cards can be arranged, we need to compute the value for 13 factorial.

13! = (1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13) = 6,227,020,800

Both 2 and 3 are clearly the diadvantages of circular list

7 does not occur in 1000. So we have to count the number of times it appears between 1 and 999. Any number between 1 and 999 can be expressed in the form of xyz where 0 < x, y, z < 9.

1. The numbers in which 7 occurs only once. e.g 7, 17, 78, 217, 743 etc

This means that 7 is one of the digits and the remaining two digits will be any of the other 9 digits (i.e 0 to 9 with the exception of 7)

You have 1*9*9 = 81 such numbers. However, 7 could appear as the first or the second or the third digit. Therefore, there will be 3*81 = 243 numbers (1-digit, 2-digits and 3- digits) in which 7 will appear only once.

In each of these numbers, 7 is written once. Therefore, 243 times.

2. The numbers in which 7 will appear twice. e.g 772 or 377 or 747 or 77

In these numbers, one of the digits is not 7 and it can be any of the 9 digits ( 0 to 9 with the exception of 7).

There will be 9 such numbers. However, this digit which is not 7 can appear in the first or second or the third place. So there are 3 * 9 = 27 such numbers.

In each of these 27 numbers, the digit 7 is written twice. Therefore, 7 is written 54 times.

3. The number in which 7 appears thrice - 777 - 1 number. 7 is written thrice in it.

Therefore, the total number of times the digit 7 is written between 1 and 999 is

243 + 54 + 3 = 300

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.

Both of them are relation ship constraints

The relationship where both the participating entities belong to same entity type is called Recursive relationship

The no of entities participating in relation is called Degree of relationship

Entity is a real world object