Combination of objects
Jun 16, 2020, 16:45 IST
Combination of objects
Definition :-
An arrangement of objects in which the order is not important is called a combination. This is different from permutation where the order matters. For example, suppose we are arranging the letters A, B and C. In a permutation, the arrangement ABC and ACB are different. But, in a combination, the arrangements ABC and ACB are the same because the order is not important.
Formula descripion :-
Mathematically, the formula for determining the number of possible arrangements by selecting only a few objects from a set with no repetition is expressed in the following way:
Where:
• N – the total number of elements in a set
• K – the number of selected objects (the order of the objects is not important)
• ! – factorial
Real life application :-
Select 11 players from 15 players team for the World Cup Final 2019.
Suppose Our Cricket team has performed tremendously in the world Cup 2019 and due to the performance they are able to qualify for the final of the world Cup. But a very gigantic problem occurs in front of selectors and captain Virat Kohli, the best spin bowler of the team injured and due to which he will not able to play for 2 weeks and the final of the World Cup is round the corner.
So Indian cricket team selectors and captain has a choice between Akshar Patel, Yujvendra chahal and Amit Mishra and they have 1 vacant position. So they can select 1 candidate out of 3 candidate in 3C1 = 3 Ways, either Akshar or Yujvendra or Amit.
Example 1 :-In how many ways can a coach choose three swimmers from among five swimmers?
Solution :-
There are 5 swimmers to be taken 3 at a time.
Using the formula:
The coach can choose the swimmers in 10 ways.
Example 2 :-Find the number of subsets of the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} having 3 elements.
Solution :-
The set given here have 10 elements. We need to form subsets of 3 elements in any order. If we select {1,2,3} as first subset then it is same as {3,2,1}. Hence, we will use the formula of combination here
Therefore, the number of subsets having 3 elements =
= 10!/(10-3)!3!
= 10.9.87!/7!.3!
= 10.9.8/3.2
= 120 ways.
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