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**Aptitude Shortcuts on Permutation and Combination**

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Before
proceeding to the problems related to the Permutation we are going to explain
little a bit about this Topic. First read carefully before solving any Problem.
Hope this will surly helps you out.

To Check Out Shortcuts on
Percentage, Profit & Loss

**CLICK HERE****This method of Permutation & combination is totally related to the math calculation and it is very important for all bank examinations.**

__Some Basic & General Rule & Formula:__**Let’s consider that “n” be a positive integer. Then, Factorial n, denoted by**

__Factorial Notation:__**⌊**

**n**or

**n!**And can be defined as follow:

n!
= n (n – 1 ) ( n –2 ) …………….3.2.1.

For Example: 5!
= ( 1 x 2 x 3 x 4 x 5 ) = 120;

For Example: 4!
= (1 x 2 x 3 x 4 ) = 24;

For Example: 3
= ( 1 x 2 x 3 ) = 6;

You should know that 1! =
1

And also the 0! = 1

Permutation:
Arrangements of a given number or any things by taking all or some at a time is
stands for permutation.

All
permutation represented by letter a, b, c by taking two at a time are ( ab, ba,
ac, ca, bc, cb )

While
All permutations represented by letters a, b, c, taking all at a time are (
abc, acb, bac, bca, cab, cba ).

**No. of all permutations of n things, taken r as a time, can be given as:**

__No. of Permutations:__

__Formula1:__

^{n}

**p**

_{r}_{ }= n (n – 1) ( n – 2 )……(n – r + 1 ) = n! / (n – r ) !

Example:

^{6}p_{2 }= 6! / 6-2= (6 x 5) = 30.
Example:

^{ }^{7 }p_{ }_{3}= (7 x 6 x 5) = 210.**No. of all arrangement that is n things, we can take all at a time = n!**

__Important Note:__
For
Example: 6! / 2! = 1 x 2 x 3 x 4 x 5 x 6 / 1 x 2 = 360.

Now
we learn what is permutation and related formula with it. Now for more practice
we are going to solve some examples of permutation with using maths shortcuts.

**Question1.**‘HOUSE’ Word can be arranged in how many Different ways?

**Solution:**As we all know that word HOUSE contains 5 letters.

Therefore required No. of word using formula is

5p5 = 5! = (1 x 2 x 3 x 4 x 5 ) = 120.

**Question2.**The words ‘COMPUTER’ can be arranged in How many different ways ?

**Sol.**We know that the word “COMPUTER” contains 7 letters. In this letter no repeated word so we can find required no. of word using formula:

^{ n}

**p**

_{r}_{ }

7p7 =
7! / ( 7 – 7)! = ( 1 x 2 x 3 x 4 x 5 x 6 x 7 ) = 5040.

**Question3.**‘KOLKATA’ word can arranged in how many different ways?

**Sol.**The Total No. of letter in this word is 7. We also w=know that repeated word occurred so we divided this by repeated word. Repeated is k two time and A two time.

Therefore
required No. of word using formula: 7p7 / 2p2 x 2p2

= 7!/(7 – 7 )! / 2!(2 – 2)! x 2!(2 – 2)! = ( 1 x 2 x 3 x 4 x 5 x 6 x 7 )

= 1260.

= 7!/(7 – 7 )! / 2!(2 – 2)! x 2!(2 – 2)! = ( 1 x 2 x 3 x 4 x 5 x 6 x 7 )

= 1260.

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