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Nov 14, 2014

MATHS SHORTCUTS ON PERMUTATION & COMBINATION PROBLEMS PART-1

Aptitude Shortcuts on Permutation and Combination
Published By: WWW.GovtJobsPortal.IN
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
Some Basic & General Rule & Formula: This method of Permutation & combination is totally related to the math calculation and it is very important for all bank examinations.
Factorial Notation: Let’s consider that “n” be a positive integer. Then, Factorial n, denoted by 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 Permutations: No. of all permutations of n things, taken r as a time, can be given as:
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.

Important Note: No. of all arrangement that is n things, we can take all at a time = n!
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.


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