Permutation and Combination Problems

Permutation is arrangement of n objects taken r at a time. You keep arranging them by taken r number of objects at a time or take n objects at a time.

Consider 4 objects –  A , B ,C, D  and 2 places 1 ,2, 3 . How many way can you arrange it?



A  B  C              B  C  D               A  B  D          A  C  D     
A  C  B              B  D  C               A  D  B          A  D  C
B  A  C              C  B  D               B  A  D          C  A  D
B  C  A              C  D  B               B  D  A          C  D  A
C  A  B              D  C  B               D  A  B          D  A  C
C  B  A              D  B  C               D  B  A          D  C  A

There is 24 ways to arrange 4 objects in 3 places.

permutation is denoted as nPr 

nPr = n!/ (n-r)! 

Let’s compute result for our previous example, 

4P3 = 4! / (4 -3)!  = 24 ways

We get same result both ways.
If we decide to arrange all n object in r places , then

n x  n x n x ….r times = nr ways

Combination is selection of n objects given r at a time or all n objects at a time. Combination is denoted as nCr .

nCr = n!/r!(n-r)!

For example, select 3 objects from 4 objects, then

4C3 =   4!/3!(4 – 3)! = 4 x 3!/ 3! . 1! = 4 ways

 Consider our previous example of permutation ,we selected one combination from each of the column. This is because in combination order of objects does not matter. Each column is combination of 3 objects only. 
But in permutation the order of object is counted as one arrangement.


A  B  C      B  C  D        A  B  D   A  C  D     = 4 ways
A  C  B              B  D  C               A  D  B          A  D  C
B  A  C              C  B  D               B  A  D          C  A  D
B  C  A              C  D  B               B  D  A          C  D  A
C  A  B              D  C  B               D  A  B          D  A  C
C  B  A              D  B  C               D  B  A          D  C  A

Problems for Math Exam:

Q1:  How many way to form a committee of 6 men and 2 women out 10 men and 5 women?

source : Mathematical foundation by p.r.vittal
 
Solution:

Number of ways to select 6 men = 10C6 

= 10! /  6! ( 10 – 6 )!

=  10 x 9 x 8 x 7 x 6! / 6 ! x  4 !
                
          3
= 10 x 9 x 8 x 7 / 4 x 3 x

=  30 x 7 
= 210 ways

Number of way to select 2 women = 5C2

= 5 !/ 2 ! ( 5 – 2)!
            
        2
= 5 x 4 x 3!/ 2! x 3!

= 10 ways


Therefore, Total ways to form the committee with 2 women and 6 men is 210 x 10 = 2100 ways.

Q2: From 6 boys and 4 girls, 5 are to be selected for admission to a particular course . In how may ways this can be done if there must be exactly 2 girls?

source : Mathematical foundation by p.r.vittal

Solution: 

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If there are exactly two girls then we must  select at least 3 boys for the course.

Number of ways to select girls = 4C2 

= 4! / 2! (4 – 2)! 

    2
= 4 x 3 x 2 ! / 2! . 2!
= 2 x 3 
= 6 ways

Number of ways to select 3 boys = 6C3 

= 6! / 3! (6 – 3)!

 
= 6 x 5 x 4 x 3! / 3! ( 3)!

= 20 ways 

Hence , Total way to  select 5 student with exactly 2 giral is 120.

Q3: How many  3 letter works can be created using letters of DEFAULT.

Solution:

Number of arrangement of 3 letters out of  7 letters of word = DEFAULT.

7P3 =  7! / (7 – 3)! 

= 7 x 6 x 5 x 4! / 4 !

= 30 x 7 

= 210 words

Now, I you can try a few 

Q 4 : How many words can be created using letters of word – EXAMINATIONS. ?

Q 5:  Find number of even numbers between 100 and 1000 ?

 
 

 
 

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