The concept of combination is used:
(a) to find number of ways element can be selected from given set, and
(b) when order is not important
Example 01
Find number of ways we can select two objects from given objects A, B, C, D
Manual Calculation
By solving the question manually we get:
AB, AC, AD, BC, BD, CD
These are the possible combinations (Total 6)
Note:
We have not included BA, CB etc. because AB = BA (both are same object & order is not important)
Solving using Combination concept
Without manual counting, use following formula:
C (n, r) = C (4, 2) = \frac{4!}{( 4-2) !\times 2!}\
C (4, 2) = 6
Same question easily solved using combination formula
Example 02
Three Badminton players are available Adam, Bill, Clark. You have to select two of them to form a team. Find the different possible combination
Manual Counting
(Adam, Bill), (Bill, Clark), (Clark, Adam)
There are three possible combination
Note: we have ignored elements like (Bill, Adam) because this will lead to double counting as:
(Adam, Bill) = (Bill, Adam)
Here order is not important
Solving using Combination formula
C (n, r) = C(3, 2) = \frac{3!}{( 3-2) !\times 2!}\
C (3, 2) = 3
Hence there are three possible combinations
Combination Definition and Formula
Combination is a math concept through which you can find the number of possible selection of element from a given set, when the order of element is not important.
Symbol & Formula of Combination
Where;
n = number of element in set
r = selected elements
Difference between Permutation & Combination
Permutation is used when order of element is important.
For example, in question where arrangement AB & BA are considered different, permutation should be used.
Combination is used when order is not important
When arrangement AB & BA are considered same, combination can be used.
Example 01
Find number of ways you can choose two elements from the set {A, B, C, D}.
Here order of elements is important [ i.e AB & BA are different]
Solution using Permutation
P (4, 2) = \frac{4!}{( 4-2) !}
P (4, 2) = 12
Hence there are 12 possible combinations.
Permutation Analysis
Below image shows how permutation works
The Permutation Formula:
1. First select the elements
2. Then reshuffle the element to gather all possible combination
3. Repeat the step 1
After all the steps we get 12 total elements as our answer
Solution using combination
Combination formula find the number of possible combination without order sequence
C (4, 2) = \frac{4!}{( 4-2) !\ 2!}
C (4, 2) = 6
Now we have to shuffle the two selected elements
=> 6 * 2!
=> 12
We got the same answer using combination formula.
Let us understand how the above process works through below illustration
Solution Explanation using Combination
1. C (4,2) gave all possible selection
AB, AC, AD, BC, CD, BD
2. The selected elements should be reshuffled to get all the possible options
AB reshuffled to give AB, BA
AC ————————-> AC, CA
AD ————————-> AD, BA
BC ————————-> AB, BA
CD ————————-> AB, BA
BD ————————-> AB, BA
For reshuffling two elements we multiplied 2!
3. Final Equation was => c (4,2) * 2!
Relationship between Permutation and Combination
In the above question we got the same answer using different technique, so there must be common link between the concept of permutation and combination.
In Permutation Method we used following equation
P (4, 2) = \frac{4!}{( 4-2) !}
In Combination Method, we used following method
=> C (4, 2) * 2!
=> \frac{4!}{( 4-2) !\ 2!} * 2!
Equating both the equation, we will get following relationship
P (n, r) = C (n, r) * r!
Note:
Just remember the following concept:
(a) Combination [C(n, r)] is only selection
(b) Permutation [P(n,r)] is both selection and rearrangement
Important Properties of Combinations
1. When r = n, the value of combination will be 1
2. When r = 0, the value of combination will be 1
3. In a collection of n objects, selecting r objects is same as rejecting (n – r) objects
4. Combination Formula
5. Combination Equality Concept
The above two combinations are equal in two condition
(1) Either a = b, or,
(2) n = a + b
All the above mentioned concepts are important.
They will help you solve worksheet problems from this chapter.
Solved Combination Problems
Question 01
A committee of 3 persons is to be constituted from a group of 2 men and 3 women. In how many ways can this be done? How many of these committees would consist of 1 man and 2 women?
Solution Part 01
Total Men = 2
Total Women = 3
Total People = 2 + 3 = 5
Let us represent all the 5 people as M1, M2, W1, W2, W3
Number of ways 3 people can be selected = C( 5, 3) = \frac{5!}{3!\ 2!}\
==> C (5, 3) = 10
Hence there are 10 ways in which three person can be selected.
Solution Part 02
Selection of 1 Man and 2 Women
Hence there are 6 possible ways in which the committee will be formed where there will be 1 Man and 2 Woman
Question 02
How many words, with or without meaning, each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE ?
Solution
Let us first separate vowels and consonant from the word INVOLUTE
Vowels: I, O, U, E [ 4 choices]
Consonant: N, V, L, T [ 4 choices]
On solving the above expression we will get = 2880 possibilities