Aptitude Shortcuts on Permutation and Combination
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.
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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.
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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.
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
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.