Important Permutations & Combinations Concept

[aneesh.kg]

I want to discuss a very important concept in Permutations & Combinations. This concept lies at the heart of this topic.

Please Read carefully.

Question: There are five people named A,B,C,D and E. In how many ways can a team of 2 people be formed from these 5 people?

Solution 1:
2 people can be selected out 5 people in 5C2 ways.
So, the number of teams = 5C2 = 10

Solution 2:
Let's select one person from the 5 people. The number of ways of doing this is 5C1.
From the remaining 4 people (i.e. excluding the person already selected), let's selected one more person. The number of ways of this is 4C1.
So, total number of ways = 5C1*4C1 = 20

What did just happen? I solved the question with two methods, both of which seem correct and yet the answers from the two methods are different. Infact, the first one is just half of the other. Can someone explain which one is correct and what is the problem with the other one?

Please don't come up with an explanation unless you are convinced of your own explanation.

Think!

[mathbyvemuri]

Aneesh it's a good one.
The answer is 5C2, ie., 10 only.
In the first method we get the following combinations: AB,AC,AD,AE,BC,BD,BE,CD,CE,DE
But,in the second method we get some additional combinations which are just repeats but in a reverse order: AB,AC,AD,AE,BA,BC,BD,BE,CA,CB,CD,CE,DA,DB,DC,DE,EA,EB,EC,ED
The underlined ones are repeats in reverse order. These are redundant as it makes no difference to have BA as a team when you already have AB at hand.
Kudos to your post, which is conceptual.

[gmatNooB8787]

I understand your reply "mathbyvemuri" , but this set thing becomes obvious only when we start searching for actual values in the sample set, but not while calculating.

[mathbyvemuri]

"gmatNooB8787", I did not get your point, pl elaborate on that

[aneesh.kg]

I don't see too many attempts on this but I hope a few people did think about this concept.

Before answering this, let me revise a few concepts of Permutations & Combinations for you pertaining to this problem.
(i) Arrangement of n different objects in n places is by n! ways
(ii) The number of ways of selecting r objects from n different objects is nCr.
(iii) AND means Multiplication.

The number of teams of 2 people from 5 people = The number of ways of selecting 2 people from 5 people = 5C2

So, Solution 1 is correct. I am sure you would've figured this much.

Solution 2 is wrong. But what is the mistake?
The mistake is in selecting objects one by one from the same pool.
Say we select A out of the 5 people and then select B.
In a separate way of selection, say we select B first, and A later.
So, we got (A,B) as well as (B,A) - which is the same team.
But, we wanted this team just once. Notice the question just wants teams of people and not their arrangement also.

We have arranged the two people of each team in 2! ways also, which is unnecessary. We have (C,D) as well as a (D,C), a (B,E) as well as a (E,B), and so on. No wonder the wrong answer is 2! times more than the correct answer of 10.

To get the correct answer, you can now divide 5C1*4C1 by 2! to nullify the redundant cases.
(5C1*4C1)/2! = 10

Do you notice now that 5C2 (or rather nCr) is such an awesome formula?
5C2 is inbuilt with a 2! at its denominator.
5C2 = (5C1*4C1)/2!
The formula makes sure that arrangements are nullified and only unique set of teams are counted. No worries!

I have seen a lot of students make this mistake. So, it's important that you understand this well.

Method3 in this solution:
https://www.beatthegmat.com/probability- ... tml#469279
is based on this concept.

@mathybyvemuri: you were correct. thanks for your input.

[sunman]

I want to discuss a very important concept in Permutations & Combinations. This concept lies at the heart of this topic.

Please Read carefully.

Question: There are five people named A,B,C,D and E. In how many ways can a team of 2 people be formed from these 5 people?

Solution 1:
2 people can be selected out 5 people in 5C2 ways.
So, the number of teams = 5C2 = 10

Solution 2:
Let's select one person from the 5 people. The number of ways of doing this is 5C1.
From the remaining 4 people (i.e. excluding the person already selected), let's selected one more person. The number of ways of this is 4C1.
So, total number of ways = 5C1*4C1 = 20

What did just happen? I solved the question with two methods, both of which seem correct and yet the answers from the two methods are different. Infact, the first one is just half of the other. Can someone explain which one is correct and what is the problem with the other one?

Please don't come up with an explanation unless you are convinced of your own explanation.

Think!

Solution 1 is correct.

Solution 2 is wrong because in a team of two, order doesn't matter. DE is the same exact team as ED, so the answer is:

(5! x 4!)/(2!) = 10.