Permutation and Combination Solved Problems

  • Permutation :
 It means arrangement where order of thing is considered.

  • Combination : 
It means selection where order of thing is not considered.

# Methods

  • Sum Rule
  • Product Rule

# Case

  • Simple 
  • Vowel comes together
  • Vowel not comes together 
Arrangement and Selection
Permutation and Combination Answer with Examples

Examples

 #1

 Arrange the word " MANISH " in following way:
 a) In how many ways the word " MANISH " can be arrange .
 b) Arrange the word " MANISH " if vowels come together.
 c) Arrange the word " MANISH " if vowel not comes together.


a)
Sol: 
MANISH
 Total no. of words = 6
 Find 6! ( Factorial = ! )
6! = 6×5×4×3×2×1
6! = 720 
Check whether there is any repeating words are there in " MANISH"
because there is no repeating words are there that's why no further action is required.
Hence Answer is 720.

b) 
Sol: 
word = " MANISH "
Step 1. Count total no. of  consonant and add 1 with it.Here, total no. of consonant = 4,now just add 1 with it .so it became (4+1=5) 5. 
Now let this outcome as factorial ( like 5!) 
Step 2. Count total no of vowels which is here 2.
Now let this outcome as factorial ( like 2!) 
Step 3. Multiply the result from Step 1 & 2. which would be like this 5! × 2! = 120 × 2 = 240.
Step 4. Check for the repeating words. Now there is no repeating words are there that's why no further action is required.
[Note : But what would you did if there were any repeating digit.To understand this.We Let word "CONDITION". Now here in this word, the repeating words are "O,N,I" then we multiply 2! upto three times (because no. of words repeating =3) and divide it with the outcome of Step 3. ]
240 is the correct answer for the word " MANISH ".

c)
Sol:
Step 1. First of all , you need to find the total no. of  possible ways in which a word can arrange without considering vowels. or Just like Sol: (a).
Step 2. Now find the total no. of possible ways of arranging a word if vowels come together .Just like Sol: (b)
Step 3. Subtract the outcome of Step 2 from Step 1.just like ( Sol (a) - Sol (b).

#2 

In how many ways  word " SUCCESS" can be arranged ?
Sol:
Total no of words = 7
Repeating words = C,S
No of times repeating = (C= 2 times & S = 3 times)
Hence 7!/3!×2! = 420


  • Arrangement of r out of n different things
  • Arrangement with repetition
  • Arrangement in a Row
  • Circular Arrangement

Arrangement of r out of n different things

permutation and combination formulas
Example : How many three-digit numbers can be formed by using the digits in 452145, if repetition is not allowed?
Sol:
Total no. of digit (n) = 6
No. of digit to be taken at a time = 3 ( required for 3 digit number).
Since repetition is not allowed, so, for making each 3- digit number, the digits chosen will be different.
hence n=6 , r=3
number of 3 -digit number formed = npr = 6p3 
= 6!/(6-3)!

Arrangement with repetition

Arrangement with repetition formula
Example: In a birthday party, Ram has 7 friends to invite.In how many ways can he send invitation cards to them if he has 4 servants to carry the cards and to deliver the same.
Sol:
Required no of ways = ( repeating thing )non-repeatable thing
Required no of ways = ( 4)7

Arrangement in a Row

Let two groups .One group has n members or persons & second has m members or persons such that n ≥ m who are sit in a row in a way that no two people or person of the same sex sits together.
No. of such sitting arrangement can be found out as :L
Permutation in row problems
Example : In how many ways can 6 boys & 4 girls will sit if no two people of the same sex are allowed to sit together .
Sol: There are two ways
(1) B G B G....so on (2) G B G B...so on.
Permutation and combination

Circular Arrangement

a) Considering Clockwise & Anti-Clockwise different
No. of possible arrangement of 'n' different things arranged around a circle is (n-1)!
b) Considered Not Different
No. of circular permutation of n different things = 1/2 (n-1)!
Example 
No. of possible ways in which 6 peoples can be arranged in a circular table .
Sol:
= (6-1)!
= 5!
Because there is no end point in circular arrangement so no of available space is reduced by 1.

Combination

Selection of r out of n different things

Theorems on Combinations
Example:
In how many ways can a cricket  team of eleven player be selected out of a batch of 15 players ?
Sol:
Required no. of ways = 15C11 = 15C(15-11) = 15C4
= 15×14×13×12/4×3×2×1
= 1365

Division into groups


Division into groups
Example :
In How many ways, Ram can select one or more of the 5 color balls from a box?
Sol:
Required no. of selection = 25 - 1 = 31


Read basic concepts of Probability here

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5 comments:

  1. check it
    nPr is used when numbers are distinct


    Example : How many three-digit numbers can be formed by using the digits in 452145, if repetition is not allowed?
    Sol:
    Total no. of digit (n) = 6
    No. of digit to be taken at a time = 3 ( required for 3 digit number).
    Since repetition is not allowed, so, for making each 3- digit number, the digits chosen will be different.
    hence n=6 , r=3
    number of 3 -digit number formed = npr = 6p3
    = 6!/(6-3)!

    ReplyDelete
  2. Sir in Ex.1(COFFEE) vowels together case u've multiplied with total no. of consonants(both are in equal numbers), but in Ex.2(CONDITIONS) u've multiplied with total no. of vowels....Only marking mistake...?

    ReplyDelete

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