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 2
= 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:
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 ?
