Combinations
Combination is a way of choosing items from a set, such as (unlike permutations) the order of selection doesn’t matter. In smaller cases, it’s possible to count the number of combinations. Combination refers to the mixture of n things taken k at a time without repetition. To know the combinations in the case where repetition is allowed, the terms like k-selection or k-combination along with repetition are often used. For instance, if we’ve two elements A and B, then there’s just one way to select two items, we select both of them. In this article, we will learn about combination in detail.
Basic Principles of Counting
Counting various things in a specific manner is a basic concern of mathematicians. To solve this problem two basic principles of counting are given which include,
- Fundamental Principle of Counting: For any event X which occurs in n different ways and another event Y which occur in m different ways. Then the total number of occurrences of two events is = m x n.
- Addition Principle: For any event X which occurs in m different ways and event Y which occurs in n different ways where both events are implicit i.e. both the events cannot occur together, then the occurrence of events either X or Y is m + n.
What Are Combinations?
Combination is the choice of selecting r things from a group of n things without replacement and where the order of selection is not important.
Number of combinations when â€˜râ€™ elements are selected out of a complete set of â€˜nâ€™ elements is denoted by ^{n}C_{r}
^{n}C_{r }= n! / [(r !) Ã— (n – r)!]
Example: Let n = 4 (E, F, G, H) and r = 2 (consisting of all the combinations of size 2).
^{n}C_{r} = ^{4}C_{2} = 4!/((4-2)!Ã—2!)
= 4Ã—3Ã—2Ã—1 / 2Ã—1Ã—2Ã—1
= 6
The six combinations are EF, EG, EH, FG, FH, and GH.
Combination Formula
Combination formula is used to pick r things out of n different things, where the order of picking is not important and replacement is not allowed.
Relationship Between Permutations and Combinations
Permutation and combination have a lot of similarities but they also have some striking differences. For n different objects, we have to make r unique selections from this group of n objects.
The number of permutations of size r from n object is ^{n}P_{r} here the order, of selection is not important so each selection is counted r! times. So the number of unique selections is ^{n}P_{r} / r! We know that a unique selection of r things from the total of n things is called a combination(^{n}C_{r}). Thus,
^{n}C_{r} = ^{n}P_{r} / r!
How to Calculate the Probability of Combinations?
The probability of Combinations can be easily understood with the help of the examples given below:
Example:
- How many ways are possible to distribute 7 different candies to 3 people where each gets only 1 candy?
- In how many ways can the letters of the word â€˜POWERâ€™ be arranged?
- How many six-digit numbers can be formed with digits 2,3,5, 6, 7, and 9 and with distinct digits?
Solution:
- For the first people, we can choose any of the candy among the 7 candies available. Similarly, for the 2nd person we are left with 6 choices and for the 3rd, we will be having 5 choices. So, the number of ways of distributing candies = 7 Ã— 6 Ã— 5 = 210 ways
- Letters of the word â€˜POWERâ€™ can be arranged in 5! ways i.e. 5 Ã— 4 Ã— 3 Ã— 2 Ã— 1 ways = 120 ways.
- The number of distinct ways of forming 6-digit numbers with different digits is 6! = 6 Ã— 5 Ã— 4 Ã— 3 Ã— 2 Ã— 1 ways = 720 ways.
What is Handshaking Problem?
Handshaking problem is one of the most interesting problems in mathematics. It is used to find that in a room full of people how many handshakes are required for everybody to shake everybody else’s hand exactly once?
Example: The table given below tells us about the minimum number of handshakes required for various groups of people.
Basically when there are 2 people there will be two handshakes and if there are three people there will be 3 handshakes and so on. This many people we can count but let’s suppose there are thousands of people in a hall then we can’t count each handshake here the need for the combination arises.
Number of People | Possible Combinations | Minimum Handshake required |
---|---|---|
Two People | A-B | 1 handshake |
Three People |
A-B A-C B-C |
3 handshake |
Four People |
A-B A-C A-D B-C B-D C-D |
6 handshake |
Handshaking Combination
It means the total number of people in a room doing the handshake with each other. With the help of combination formulas, it can easily be calculated. The formula for calculating the handshakes when there are n people available is given by,
- Total Number of Handshakes = n Ã— (n – 1)/2
- Total Number of Handshakes = ^{n}C_{2}
Also, Check
Solved Examples on Combinations
Example 1: In how many ways 6 boys can be arranged in a queue such that
a) Two particular boys of them are always together
b) Two particular boys of them are never together
Solution:
a) If two boys are always together, then they will be treated as one entity. Hence we can be arranged 5 boys in 5! ways. Also, two boys can arrange themselves in 2 different ways.
Therefore required arrangement = 5! Ã— 2 = 120 Ã— 2 = 240 ways.
b) Total number of permutations among 6 numbers is given by = 6! = 720.
In 240 cases 2 boys are always together.
Thus, for two boys who are never together no of ways will be = 720 â€“ 240 = 480 ways.
Example 2: In a room of n people, how many handshakes are possible?
Solution:
To see the people present, and consider one person at a time. The first person will shake hands with n – 1 other people. The next person will shake hands with n-2 other people, not counting the first person again. Following this, it will give us a total number of
(n – 1) + (n – 2) + … + 2 + 1
= n(n – 1)/ 2 handshakes.
Example 3: Another popular handshake problem starts out similarly with n>1 people at a party. Not being possible to shake hands with yourself, and not counting several times handshakes with the same person, the problem is to show that there will always present two people at the party, who had shaken hands the same number of times in the party.
Solution:
The solution to this problem starts by using Dirichlet’s box principle. If there exists a person at the party, who has shaken hands zero times, then every person which is there at the party has shaken hands with at most n-2 other people at the party.
There are n-1 possible handshakes (from 0 to n-2), among n people there must be two who have shaken hands the same number of times. If there are zero persons, who has shaken hands zero times this means that all of the party guests have shaken hands at least once.
This also amounts to n-1 possible handshakes (from 1 to n-1).
Example 4: In the function, if every person shakes hands with every other in the party and there exists a total of 28 handshakes at the party, find the number of persons who were present in the function.
Solution:
Suppose there are n persons present at a party and every person shakes hands with every other person.
Then, total number of handshakes = ^{n}C_{2 }= n(n – 1)/2
n(n – 1)/2 = 28
n(n – 1) = 28 Ã— 2
n(n – 1) = 56
n = 8
FAQs on Combination
Question 1: What is a Combination?
Answer:
The combination is a way of arranging r different things out of n things for which the order of selection is not important.
Question 2: How to solve combinations?
Answer:
Combinations help us to calculate the total outcomes of an event when the order of outcomes does not matter. Combinations can be calculated with the formula,
^{n}C_{r} = n! / r! Ã— (n – r)!
Question 3: What is the value of ^{n}C_{n}?
Answer:
The value of^{ n}C_{n} is calculated as,
^{n}C_{n }= n! / (n-n)!Ã—n! (0! = 1)
= n! / n! = 1
Question 4: When do we use Combination and Permutation?
Answer:
Permutation formulas are used when the order of selection matters and Combination formulas are used when the order of the permutation doesnâ€™t matter.
Question 5: Give the Combination formula.
Answer:
Combination formula is given as,
^{n}C_{r} = n!/r!(n-r)!
Question 6: What do you mean by Derangement?
Answer:
When we shuffle the elements of a set so that no element appears in its original position it is called derangement of data.
Please Login to comment...