__A-I) To find the Rank of a word__
This
is one of the traditional types. Here we elaborate on a simple example to find
rank of word ‘MATH’:

First
we arrange the letters alphabetically - A,H,M,T

Now
we find number of words starting with each letter in the alphabetical order.

Number
of words starting with A = 3! = 6 (Arrangements
of remaining three letters – ‘M’ , ‘T’ and ‘H’)

Number
of words starting with H = 3! = 6 (Arrangements
of remaining three letters – ‘A’ , ‘M’ and ‘T’)

Our
word ‘MATH’ is in the next set of words ie., words starting with ‘M’

First
word in this set- MAHT

Second
word is –

**MATH,**there we are. So rank of the word = 6 + 6 + 2 = 14

__A-2) To find the number of straight lines passing through distinct points__
Given
‘m’ points in a plane out of which ‘p’ points are collinear and no other three
points are collinear.

To
form a line two points are required. So from the given ‘m’ points, we are
supposed to get mC2 lines (why not mP2 ???,because we are selecting two points
and thats all, we need not arrange them). But since

‘p’ out of the given ‘m’ points are collinear, by joining these ‘p’ points we get only one line instead of pC2 lines.

‘p’ out of the given ‘m’ points are collinear, by joining these ‘p’ points we get only one line instead of pC2 lines.

The
number of lines formed = mC2 – pC2 + 1

__A-3) To find the number of triangles formed by joining distinct points__
Given
‘m’ points in a plane out of which ‘p’ points are collinear and no other three
points are collinear.

**To form a triangle three points are required**. So from the given m points, we are supposed to get

**mC3**triangles. But since ‘p’ out of the given ‘m’ points are collinear, no triangle can be formed from these ‘p’ points instead of estimated pC3 triangles.

**The number of triangles formed = mC3 –pC3**

__A-4) Two sets of intersecting parallel lines forming parallelograms__
Find
the number of parallelograms formed when a set of m parallel lines intersects
another set of n parallel lines:

Parallelogram
is formed when one pair of parallel lines intersects with another pair. Hence
we need to select one pair each from the given two sets of parallel lines. Here
selection only is involved but no arrangement. So we go for nCr but not nPr.

One
pair of lines (ie., 2 lines) can be selected from first set of ‘m’ lines in mC2
ways and second pair from second set of ‘n’ lines in nC2 ways. Both can be
performed in (mC2)*(nC2) ways (can be understood form the fundamental principle).

**Hence number of parallelograms formed = (mC2)*(nC2)**

__A-5) Number of Squares and Rectangles in a nXn square__

Consider
the following 2X2 square:

To
find number of squares in the above diagram:

Number
of 1X1 squares = 4 (ABED, BCFE, DEHG,
EFIH)

Number
of 2X2 squares = 1 (ACIG)

Total
number of squares = 4 + 1 = 5 = 2

Similarly for 3X3 square, we get 14 = 9+4+1 = 3

The general formula is

^{2}+ 1Similarly for 3X3 square, we get 14 = 9+4+1 = 3

^{2}+2^{2}+1The general formula is

**Σn**^{2}
To
find number of rectangles in the above diagram:

(We
consider squares also as rectangles)

Number
of 1X1 rectangles = 4 (ABED, BCFE, DEHG, EFIH)

Number
of 1X2 and 2X1 rectangles = 4 (ACFD,
DFIG, ABHG, BCIH)

Number
of 2X2 rectangles =1 (ACIG)

Total
number of rectangles = 4 + 4 + 1 = 9 which can be expressed as

Similarly for 3X3 square, we get 36 , which can be expressed as 3

The general formula for rectangles is ^{3}+2^{3}+1^{3}**Σn**

^{3}
If we consider 3X3 square, we can find 36 rectangles.

The logic here is nXn square is formed by two sets of n+1 number of parallel lines, and to form a rectangle we need 2 lines from each set thus forming (n+1)C2 * (n+1)C2 rectangles

**Number of squares in a nXn square =**

**Σn**

^{2}**Number of rectangles in a nXn square =**

**Σn**

^{3}

__A-6) To find Number of divisors__
If
a given number n can be expressed as p

_{1}^{k1}. p_{2}^{k2}. P_{3}^{k3}...... p_{n}^{kn}
Where
p

_{1}, p_{2}, p_{3},... p_{n}are prime numbers and k_{1},k_{2},k_{3},...k_{n}are positive integers
(This
is nothing but factorisation of the given number and expressing in the form of
exponents),

**then total number of divisors of the given number = (k**

_{1}+1) (k_{2}+1) (k_{3}+1)... (k_{n}+1)
Explanation:

Observe
that this is corollary of the theorem of all possible selections (similar
things).

Divisors
of p

_{1}^{k1 }are 1,p_{1}, p_{1}^{2},p_{1}^{3},.... p_{1}^{k1}– resulting in (k_{1}+1) numbers
Divisors
of p

_{2}^{k2 }are 1,p_{2}, p_{2}^{2},p_{2}^{3},.... p_{2}^{k2}– resulting in (k_{2}+1) numbers
And so on and so on...

Divisors of p

By applying fundamental
principle, this results in _{n}^{kn }are 1,p_{n}, p_{n}^{2},p_{n}^{3},.... p_{n}^{kn}– resulting in (k_{n}+1) numbers**(k**number of divisors

_{1}+1) (k_{2}+1) (k_{3}+1)... (k_{n}+1)
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