3 Tips in Permutations and Combinations PDF

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CONTENTS Article 1: Learn when to “Add” and “Multiply in Permutation and Combination Questions Article 2: Fool-proof method to Differentiate between Permutation & Combination Questions Article 3: 3 Deadly mistakes you must avoid in Permutation and Combination

Learn when to “Add” and “Multiply” in Permutation & Combination questions

Why should you read this article? • Are you someone who is confused whether to add or multiply the cases while solving permutation and combination questions? • Do you often get questions incorrect simply because you added entities instead of multiplying them? If the answer to any of the above questions is yes, then in this e-GMAT article you will learn how to get rid of the confusion by using “AND” & “OR” on these questions.

What will you learn from this article? In this e-GMAT article, we will: 1. Dive into the details of the application of “AND” & “OR” in permutation and combination. 2. Discuss the attributes that AND – OR present through a few GMAT like questions 3. Help you to solidify the understanding of the above two keywords and at the end of the article, we will also provide you a few practice questions to test your understanding.

When to Add – Usage of the keyword “OR” While solving any P and C question, the most frequent confusion that students have is: “Do I need to add up all the cases or do I need to multiply all the cases?” Let’s understand what I am trying to say with the help of two questions! • Amy has 3 different types of shoes and 2 different types of sandals. Whenever she goes out, she likes to wear either a shoe or a sandal. In how many ways can she decide what to wear? • Joe has 3 different types of shirts and 2 different types of trousers. Whenever he goes out, he likes to wear a shirt and a trouser. In how many ways can he decide what to wear? Look at these two situations. In both these cases, the numbers are the same. The only difference is that Amy will choose either shoes OR sandals to wear and in the second case Joe is going to choose both a shirt AND a trouser.

Let us focus on Amy’s situation first, and then we will come to Joe. Assuming that she has red, green, and black shoes, and she has brown and blue sandals, let us list down all the possible options that she has. 1. She can wear Red Shoes OR 2. She can wear Green shoes OR 3. She can wear Black Shoes OR 4. She can wear Brown Sandals OR 5. She can wear Blue Sandals

Notice how we have used the word OR after each and every case. The OR here emphasizes on the fact that Amy does not have the option to wear two different kinds of things at the same time! She needs to choose only 1 of these. Thus, the answer, in this case, will be 5. Now let us use the keyword approach (OR) to solve this question easily!

Keyword Approach by e-GMAT Look at the question once again: • Amy has 3 different types of shoes and 2 different types of sandals. Whenever she goes out, she likes to wear either a shoe OR a sandal. In how many ways can she decide what to wear? o Notice the OR that I have highlighted in this question. • Whenever you read a question, look for the keyword OR or AND. While looking for the keyword, focus on what we need to find out.

o In this case, we need to find the number of ways in which she can wear a shoe OR a sandal. o Since the word OR is used here, we can conclude that both the events cannot happen together.  By event I mean, Amy cannot wear Shoe, and Sandal together. Whenever we have a situation in which two events cannot occur simultaneously, we simply add all the cases. Thus, in this case, we will say: •Total possible cases for Amy = She will wear shoes OR she will wear Sandals oNumber of ways she can wear shoes = 3 oNumber of ways she can wear Sandals = 2 •Therefore, total possible cases = 3 OR 2 = 3 + 2 = 5

Takeaway by e-GMAT experts 1- Whenever we come across a situation involving 2 or more events, and occurrence of one event does not affect the occurrence of the other event, i.e., both of the events cannot occur simultaneously, then, in that case, we will simply add up all the events! 2- Look for the word OR in the question while figuring out what you need to find out and if OR is present then that means you need to add up the events.

When to Multiply – Usage of Keyword “AND” Now let us look at Joe’s case. Assuming that he has red, green, and a black shirt, and he has brown and blue trousers, let us list down all the possible options that he has. 1. He can wear Red Shirt AND Brown Trousers 2. He can wear Red Shirt AND Blue Trousers 3. He can wear Green shirt AND Brown Trousers 4. He can wear Green shirt AND Blue Trousers 5. He can wear Black shirt AND Brown Trousers 6. He can wear Black shirt AND Blue Trousers

Notice how I have used the word AND after each and every case. The AND here emphasizes on the fact that Joe has to wear both shirt AND trousers at the same time! We cannot imagine a scenario, where

he wears just a shirt OR just a trouser.

From the above cases, we can see that he has 6 different options for wearing a shirt and a trouser. Now let use the keyword approach (AND) to solve this question easily! ?

Let’s use the Keyword “AND” Look at the question once again: • Joe has 3 different types of shirts and 2 different types of trousers. Whenever he goes out, he likes to wear a shirt AND a trouser. In how many ways can he decide what to wear? o Notice the AND that I have highlighted in this question. • In this case also focus on what we need to find out. o In this case, we need to find the number of ways he can wear a shirt AND a trouser. o Since the word AND is used here, we can conclude that both the events need to happen together.  Which means he needs to wear the shirt and trouser together. Whenever we have a situation in which two events can happen simultaneously, we simply “multiply” all the cases. Thus, in this case, we can say:

•Total possible cases for Joe = He will wear a Shirt AND he will wear Trousers oNumber of ways he can wear a shirt = 3 ( Red OR Green OR Black) [ Keyword OR: hence addition] oNumber of ways he can wear trousers = 2 (Blue OR Brown) •Therefore, total possible cases = 3 AND 2 = 3 x 2 = 6

Takeaway by e-GMAT experts 1- Please remember that whenever we come across a situation involving 2 or more events and each event can happen simultaneously, i.e., event 1, event 2, event 3 and so on, all can happen simultaneously. Then, in that case, we will simply MULTIPLY up all the events! 2- Look for the word AND in the question while figuring out what you need to find out, and if AND is present, then that means you need to multiply the events.

What to do when AND or OR keyword is not present? Now, there could be a situation when the word AND or OR is not explicitly mentioned in the question. What should we do then? Let us understand this with the help of 3 examples:

e-GMAT Example 1

Q-In the above diagram, in how many ways can you go from Point A to point D? Solution

In the question and diagram, you can see that there is no mention of the word AND or OR, so what should we do? • Well, we will simply look at the each and every end point and figure out what are we exactly doing. • So, think, in how many ways can we go from point A to B? o We can go in one way, right? o So, let us just write A -> B : 1 way • Now, when we want to go from B to C, we can go via X OR Y. o Notice how we are using the word OR here. o We know that we cannot simultaneously go via both the routes. And thus, we need to use the word OR here.  Thus, total number of ways to go from B to C = 2  B ->X ->C OR B->Y->C = 2 ways • Lastly, we want to go to our final destination D, we can go via P OR Q o In this case also, we are using the word OR. o Hence, we will simply ADD all the cases and get C->D = 2 ways. • So, now we know that we can go from A to B in 1 way, then from B to C in 2 ways and from C to D in 2 ways. o Now, ask yourself, should we add these cases or multiply? o Can we reach from A to D by going through A to B or B to C or C to D?  No, we cannot, right?  We need to go from A to B AND then from B to C AND finally from C to D to reach our destination.

• Thus, total cases possible are = 1 x 2 x 2 = 4 ways

e-GMAT Example 2

Q-How many 3 digits even numbers can be created using the digits 1,2,3, 4 and 5 without repetition of digits?

Solution

This question does not mention the keywords, AND or OR. Hence, we will try to figure out what the question is asking. Any 3-digit number can be represented as:

• We have to make a 3-digit number. o Now think, can we make a three-digit number by just filling the Units place OR the tens place OR the hundreds place?  No, we cannot, Right? o Thus, we have to fill the units place AND the tens place AND the hundreds place to get the three-digit number. o Notice, we used the keyword AND here. Hence, we will multiply all the cases

 Thus,

• We have to form a 3 digit even number. o Hence, the units digit of the number must be even. o Among the given digits: 1, 2, 3, 4, and 5, the units can be filled with 2 digits.  Either 2 OR 4.  Can you see the usage of the keyword OR here? o Hence, the number of ways to fill the units place= 2

• Now, we have filled the units digit to take care of the even nature of the number. o We have 4 digits remaining – 1 even and 3 odd. o The tens place can be filled with either 1 OR 3 OR 5 OR the remaining number between (2/4).  Thus, total ways to fill tens place= 4 ways. • Now, we only have to fill the last digit and we have 3 digits remaining. o Are you now able to visualise that we will use OR to select 1 among the 3 digits?  We can select either the 1st remaining digit OR the 2nd remaining digit OR the 3rd remaining digit  We used the keyword OR here. Hence, we will add the cases. • Thus, total ways to fill tens place= 3 ways. • As discussed, we need to multiply all these 3 individual cases to get the total number of ways in which the number can be formed. o Hence total ways= 3*4*2=24 ways

Let us come to the last question of this e-GMAT article.

e-GMAT Example 3

Q- A committee of 2 members is to be formed from a panel of 4 male and 3 female members of a club. In how many ways can the committee be formed?

Solution You should now start thinking a step ahead to find the use of AND-/OR in such problems. Let us find out how we can form the committee of 2 members in the above question.

• We can select either 2 males OR 2 females OR 1 male AND 1 female to form the committee. o Look for the keywords- AND and OR in the above statement. o Now it is easy, we know that we add the number of cases when we have an OR, and, we multiply the cases when we have an AND. o Total ways= ways to select 2 males + ways to select 2 females + ways to select 1 male * ways to select 1 female Let us find in how many we can select 2 males, 2 females and 1 male and 1 female. • We have 4 males: o We can select 2 males from 4 males in 4c2 ways and, o We can select 1 male from 4 males in 4c1 ways. • We have 3 females: o We can select 2 females in 3c2 ways and, o We can select 1 females in 3c1 ways.

Thus, total ways= 4c2+ 3c2+ 4c1*3c1 = 6+3+4*3=21 Thus, the committee can be formed in 21 ways.

Takeaway from this article 1. Specific keywords can be the X-factor in solving PNC questions. You can trust them, but at the same time apply common sense just to be sure you are on the right path. 2. Always try to figure out after reading the question whether the events are dependent or independent. Making this inference will help you solve the question more efficiently. 3. Whenever we have a situation where OR is involved, please ADD up all the events. 4. Whenever we have a situation where AND is involved, please MULTIPLY all the

events, 5. Whenever the keyword is not available, always try to jot down all the events first and then decide whether you need them together or independently.

Fool-proof method to Differentiate between Permutation & Combination Questions

Highlights of the previous article In the previous article, we discussed: • When to add and multiply by the keyword approach in Permutation and combination question. • How to solve the permutation and combination question when keywords are not present.

You can read the previous article here: Learn when to “Add” and “Multiply” in Permutation & Combination questions With the basic understanding of AND-OR keywords, let us dive into the advanced concept ofcombination and permutation. Then we will apply the learnings to solve few GMAT-like questions.

Agenda of the article In this article, we will discuss: • How to figure out when to apply combination and when to apply permutation. • Keyword approach to identify the combination or permutation type of questions. • How to visualize a permutation and combination question if keywords are not given. We will also provide few GMAT like practice questions to test the understanding.

A general case In most of the p and c questions, we arrive at a point where we need to select or arrange a few things and many students fall prey to the same mistake of applying selection in place of arrangement and vice-versa. To clarify this confusion, let us understand two simple cases: 1. From 3 players, A, B, and C, how many doubles team can be formed? 2. From 3 letters, A, B, and C, how many 2-digit words can be formed? Do both the examples looks same to you??? Well, the examples are not same. • In example 1, the team (A B) is same as the team (B A). • While, in example 2, the word AB is not same as word BA. o Thus, in the first case, arrangement of the “team members” does not affect the team composition. But, in the second case, the arrangement of the letters can give us two different words.

This simple example clearly shows that the understanding of combination and permutation can help to decide when arrangement matters and when selection matters.

Combination Let us understand the concept of combination by solving example 1- “From 3 players, A, B, and C, how many doubles team can be formed?” From 3 players A, B, and C, the teams of 2-players can be:

• Team AB • Team AC • Team BC Thus, we can have only 3 doubles teams from 3 players. Now, instead of solving this manually, let us apply the keyword approach to solve this question.

Keyword approach Let us list all the cases in which a doubles team can be formed. • Select A and select B • Select A and select C • Select B and select C Can you notice the keyword- “SELECT”, in all the cases?? In all the above cases, the selection of 2 players is same as the combination of 2 players only. Therefore, we can infer that the keyword select is used for a combination question. Now, per our understanding, the formula to select ‘r’ things from ‘n’ things, is nCrnCr, which is equal to n![(n−r)!∗r!]n![(n−r)!∗r!]. Thus, going by the above formula, we can conclude that the number of ways to select 2 players from 3 players is 3C23C2= 3![(3−2)!∗2!]3![(3−2)!∗2!] = 3 ways. Can you see that getting answer by nCrnCr formula is actually easier than manually counting all the cases??

Let us look at some frequently used keywords that imply a combination question.

Important keywords to identify a combination question Some of the important keywords are: • Select • Choose • Pick • Combination Whenever you read a question, look for the above keywords as these are the useful indicators that clearly tells us that the question is a combination question.

Let us see the application of the above keywords in 2 practice questions.

e-GMAT Example 1

Q--In a society of 10 members, we have to select a committee of 4 members. As the owner of the society, John, is already a member of the committee. In how many ways the committee can be formed.

Solution Notice the keyword- SELECT in the question. Thus, this is a combination question. And for selection, we apply the nCrnCr formula to arrive at the answer. • Now, in the question, we are asked to select a committee of 4 members from 10 members and John is already a part of the committee. o Thus, we have to select 3 members among 9 members.

• By the application of nCrnCr formula, we can select 3 members from 9 members in 9C39C3 ways which is equal to 9!6!∗3!9!6!∗3! = 84 ways

Now, let us solve a slightly difficult question.

e-GMAT Example 2

Q--An analyst will recommend a combination of 3 industrial stocks, 2 transportation stocks, and 2 utility stocks. If the analyst can choose from 5 industrial stocks, 4 transportation stocks, and 3 utility stocks, how many different combinations of 7 stocks are possible?

Solution Notice the highlighted keywords- CHOOSE and COMBINATIIONS. Now, we can easily identify that this is selection question, right?? The analyst needs to form different combination of 7 different stocks. Can you visualize how can he do that? Approach: • He needs to select 3 industrial stocks out of 5 industrial stock AND, • He needs to select 2 transportation stocks out of 4 transportation stocks AND, • He needs to select 2 utility stocks out of 3 utility stocks. o Notice the keyword ‘AND’ which indicates all the above 3 events have to occur simultaneously. o Thus,

By the application of nCrnCr formula, we can write: • 3 industrial stocks out of 5 industrial stock can be selected in 5C35C3=10 ways. • 2 transportation stocks out of 4 transportation stocks can be selected in 4C24C2=6 ways. • 2 utility stocks out of 3 utility stocks can be selected in3C23C2=3 ways. Thus, the total ways to select 7 stocks = 10*6*3 =180 ways.

Key Takeaways 1- Keep an eye on the important keywords like- select, choose, combination in the question stem. 2- The number of ways to select ‘r’ things among ‘n’ things = nCrnCr. Now, let us see how permutation works.

Permutation Let us understand the concept of permutation by solving example 2-“From 3 letters, A, B, and C, how many 2-letter words can be formed?” The 2-letter words that can be formed from 3 letters A, B, and C are: • AB • BA • AC • CA • BC • CB Thus, we can form 6 different words. Can you observe that in combination, the selection of A and B gives only 1 team i.e. AB? However, the selection of A and B gives 2 different words i.e. AB and BA. This happens because the order of arrangement in case of words matters. But while creating teams, the team composition does not change whether we say AB or BA. This arrangement is known as permutation. Can you notice the usage of keyword- ARRANGEMENT in permutation?? • If not, then keep note: The word arrangement in a question implies a permutation question. Now, instead of solving this manually, let us apply the keyword approach to solve this question.

Keyword approach Let us form all the cases in a different way. In this way, we will first select the two letters and then we will arrange the selected letters. • Select A and Select B o We now have two letters, A and B and we can arrange them in two different ways.  A then B  B then A • Select A and Select C o We now have two letters, A and C and we can arrange them in two different ways.  A then C  C then A • Select B and Select C

o We now have two letters, B and C and we can arrange them in two different ways.  B then C  C then B By counting all the cases, total 2 letter words= 6 Per our understanding, the formula to arrange ‘r’ things from ‘n’ things, is nPrnPr which is equal to n!(n−r)!n!(n−r)!. Thus, going by the above formula, we can conclude that different 2-letter word= 3P23P2= 3!(3−2)!3!(3−2)!= 6 words.

Interesting fact: • From the above example, can you see that permutation is same as doing selection first AND then doing arrangement?? (Notice- keyword: AND) • Let us understand this mathematically o nPrnPr = n!(n−r)!n!(n−r)!= n!(n−r)!∗r!n!(n−r)!∗r! * r!= nCrnCr* r...