Section 5.5: Counting Techniques1. 5.1 Probability Rules2. 5.2 The Addition Rule and Complements3. 5.3 Independence and the Multiplication Rule4. 5.4 Conditional Probability and the General Multiplication Rule5. 5.5 Counting Techniques6. 5.6 Putting It Together: ProbabilityObjectivesBy the end of this lesson, you will be able to...1. solve counting problems using the Multiplication Rule2. solve counting problems using permutations3. solve counting problems using combinations4. solve counting problems involving permutations with non-distinct items5. compute probabilities involving permutations and combinationsDo you remember the classical method for calculating probabilities from Section 5.1?P(E) =number of ways E can occur=N(E)

total number of possible outcomesN(S)

Well, sometimes counting the "number of ways E can occur" or the "total number of possible outcomes" can be fairly complicated. In this section, we'll learn several counting techniques, which will help us calculate some of the more complicated probabilities.The Multiplication Rule of CountingLet's suppose you're preparing for a wedding, and you need to pick out tuxedos for the groomsmen. Men's Tuxedo Warehouse has aBuild-A-Tuxfeature which allows you to look at certain combinations and build your tuxedo online. Let's suppose you have the components narrowed down to two jackets, two vest and tie combinations, and three shirt colors. How many total combinations might there be?A good way to help understand this type of situation is something called atree diagram. We begin with the jacket choices, and then each jacket "branches" out into the two vest and tie combinations, and then each of those then "branches" out into the three shirt combinations. It might look something like this:

In total, it looks like we have 12 possible combinations of jackets, vests, and shirts. (Of course, some may not fit your fashion sense, but that's another question all-together...)Isn't there an easier way to do this? Why yes, there is! Think of it this way, foreachjacket choice, there are two vest and tie choices. That gives us 4 total jacket and vest/tie combinations. Then, foreach of those, there are three shirt choices, giving us a total of 12.In general, wemultiplythe number of ways to make each choice, so...total number of outfits = (number of jackets)(number of vest/ties)(number of shirts)That leads us to the Multiplication Rule of Counting:Multiplication Rule of CountingIf a task consists of a sequence of choices in which there arepways to make the first choice,qways to make the second, etc., then the task can be done inpqr...different ways.Let's try some examples.Example 1How many 7-character license plates are possible if the first three characters must be letters, the last four must be digits 0-9, and repeated characters are allowed?[ reveal answer ]The total number of license plates would be:(# choices for 1st character)(# choices for 2nd)etc..= 26262610101010 = 175,760,000Example 2

Source:SearsMany garage doors have remote-access keypads outside the door. Let's suppose a thief approaches a particular garage and notices that four particular numbers are well-used. If we assume the code uses all four numbers exactly once, how many 4-digit codes does the thief have to try?[ reveal answer ]Not very many!total # of codes = (# choices for 1st digit)(# for 2nd)etc...= 4321 = 24Notice that the number of choices decreased by one for each digit, since the four numbers were used only once. You'll often see this described either as the number are chosen "without replacement" or that "repeats are not allowed".Example 2, from earlier this section is an example of particular counting technique called apermutation. Rather than giving you formulas and examples myself, I'd like to make another reference to some content from one of my favorite web sites,BetterExplained. Here's what the author, Kalid Azad writes about permutations:Permutations: The hairy detailsLets start with permutations, orall possible waysof doing something. Were using the fancy-pants term permutation, so were going to care about every last detail, including the order of items. Lets say we have 8 people:1. Alice2. Bob3. Charlie4. David5. Eve6. Frank7. George8. HoratioHow many ways can we pick a Gold, Silver, and Bronze medal for Best friend in the world?

Were going to use permutations since the order we hand out these medals matter. Heres how it breaks down: Gold medal: 8 choices: A B C D E F G H (Clever how I made the names match up with letters, eh?). Lets say A wins the Gold. Silver medal: 7 choices: B C D E F G H. Lets say B wins the silver. Bronze medal: 6 choices: C D E F G H. Lets say C wins the bronze.We picked certain people to win, but the details dont matter: we had 8 choices at first, then 7, then 6. The total number of options was 8 * 7 * 6 = 336.Lets look at the details. We had to order 3 people out of 8. To do this, we started with all options (8) then took them away one at a time (7, then 6) until we ran out of medals.We know the factorial is:Unfortunately, that does too much! We only want 8 * 7 * 6. How can we stop the factorial at 5?This is where permutations get cool: notice how we want to get rid of 5*4*3*2*1. Whats another name for this? 5 factorial!So, if we do 8!/5! we get:

And why did we use the number 5? Because it was left over after we picked 3 medals from 8. So, a better way to write this would be:

where 8!/(8-3)! is just a fancy way of saying Use the first 3 numbers of 8!. If we havenitems total and want to pickkin a certain order, we get:just means Use the first k numbers of n!And this is the fancy permutation formula: You havenitems and want to find the number of wayskitems can be ordered:

Source:BetterExplained, Kalid AzadArticle:Easy Permutations and CombinationsUsed with permission.As a side note, your text uses the notationnPkrather than Kalid's P(n,k). I've seen both used, though the later tends to be more prevalent in higher-level math classes. We'll stick with the textbook version, just to be consistent.Permutations ofnDistinct Objects Takenrat a TimeThe number of arrangements ofrobjects chosen fromnobjects in which1. thenobjects are distinct,2. repeats are not allowed,3. order matters,is given by the formula.OK, let's try a couple.Example 3Suppose an organization elects its officers from a board of trustees. If there are 30 trustees, how many possible ways could the board elect a president, vice-president, secretary, and treasurer?[ reveal answer ]In this example, we have 30 "items" (trustees), from which we're choosing 4. Using the notation from your text, we want to calculate30P4, or30P4= 30!=30!= 30292827 = 657,720

(30-4)!26!

Example 4Suppose you're given a list of 100 desserts and asked to rank your top 3. How many possible "top 3" lists are there?[ reveal answer ]100P3= 100!=100!= 1009998 = 970,200

(100-3)!97!

In the previous page, we talked about the number of ways to choosekobjects fromnif the ordered mattered - like giving medals, electing officers, or pickling favorite desserts. What if order doesn't matter, like picking members of a committee?Again, I'll let Kalid Azad explain.Combinations, Ho!Combinations are easy going. Order doesnt matter. You can mix it up and it looks the same. Lets say Im a cheapskate and cant afford separate Gold, Silver and Bronze medals. In fact, I can only afford empty tin cans.How many ways can I give 3 tin cans to 8 people?Well, in this case, the order we pick people doesnt matter. If I give a can to Alice, Bob and then Charlie, its the same as giving to Charlie, Alice and then Bob. Either way, theyre going to be equally disappointed.This raises and interesting point weve got some redundancies here. Alice Bob Charlie = Charlie Bob Alice. For a moment, lets just figure out how many ways we can rearrange 3 people.Well, we have 3 choices for the first person, 2 for the second, and only 1 for the last. So we have 3 * 2 * 1 ways to re-arrange 3 people.Wait a minute this is looking a bit like a permutation! You tricked me!Indeed I did. If you have N people and you want to know how many arrangements there are forallof them, its just N factorial or N!So, if we have 3 tin cans to give away, there are 3! or 6 variations for every choice we pick. If we want to figure out how many combinations we have, we justcreate all the permutations and divide by all the redundancies. In our case, we get 336 permutations (from above), and we divide by the 6 redundancies for each permutation and get 336/6 = 56.The general formula is

which means Find all the ways to pick k people from n, and divide by the k! variants. Writing this out, we get ourcombination formula, or the number of ways to combine k items from a set of n:

Source:BetterExplained, Kalid AzadArticle:Easy Permutations and CombinationsUsed with permission.As a side note, your text uses the notationnCkrather than Kalid's C(n,k). As with permutations, we'll stick with the textbook version, just to be consistent.Combinations ofnDistinct Objects Takenrat a TimeThe number of arrangements ofnobjects usingrnof them, in which1. thenobjects are distinct,2. repeats are not allowed,3. order does not matter,is given by the formula.All right, let's try this new one out.Example 5Let's consider again the board of trustees with 30 members. In how many ways could the board elect four members for the finance committee?[ reveal answer ]In this example, we have 30 "items" (trustees), from which we're choosing 4. Unlike in Example 3, order doesn't matter for this example, so we're looking at a combination rather than a permutation.30C4= 30!=30!=30292827 = 27,405

(30-4)!4!26!4!4321

You may notice that this number is quite a bit smaller than in Example 3. The reason is that we don't care about order now, so electing trustees A, B, C, and D for the committee is no different than electing trustees B, C, A, and D. That's different than in Example 3, where we were electing them to particular positions.Example 6

Source:stock.xchngSuppose you're a volleyball tournament organizer. There are 10 teams signed up for the tournament, and it seems like a good idea for each team to play every other team in a "round robin" setting, before advancing to the playoffs. How many games are possible if each team plays every other team once?[ reveal answer ]This may not initially seem like a combination, but let's take closer look. We have 10 "items" (teams) from which we're choosing 2. We don't care about order, since team A playing team B is the same as team B playing team A. That's exactly a combination!10C2= 10!=10!=109 = 45

(10-2)!2!8!2!21

Wow, that's a lot of games! That's why most tournaments go with a "pool" structure and split the tournament up into two "pools" of 5.You can see more on the "round robin" structure for tournaments atWikipedia.This second type is less common. What if we want to know how many ways to ordernobjects, but they're not all distinct? Here's an example to illustrate:Example 7In how many ways could the letters in the word STATISTICS be rearranged?The answer is a little tricky. Think of the rearranged words as places for letters to go. Something like this:

In STATISTICS, we have the following letters:3 S's3 T's2 I's1 A1 CWe can't really say that there are 4 choices for the first letter and proceed from there, since the number of choices for the second letter depend on which letter was chosen for the first.Instead, wechoose the spotsfor each of the letters. First, pick 3 of the 10 spots for the S's. We can do that in10C3ways. Then pick 3 spots for the 3 T's. We can do that in7C3ways. Similarly, we can pick the spots for the I's, the A, and the C in4C2,2C1, and1C1ways, respectively. In total, that means we rearrange the letters in:10C37C34C22C11C1waysIt may just be me, that isreallymessy. Oddly enough, writing out the combinations reveals a nice way to simplify it:10!7!4!2!1!=10!

7!3!4!3!2!2!1!1!1!0!3!3!2!1!1!

Permutations with Non-distinct ItemsThe number of permutations ofnobjects, where there are n1 of the 1st type, n2 of the 2nd type, etc, is

One quick example:Example 8In how many ways can the word REARRANGE be rearranged?[ reveal answer ]Wittiness aside, this is the same as the previous example. In this case, we have 3 R's, 2 A's, 2 E's, 1 N and 1 G. The total number of permutations is thus:9!= 15,120

3!2!2!1!1!

Fundamental Principles of Counting

Here we shall discuss two fundamental principles viz. principle of addition and principle of multiplication.These two principles will enable us to understand Permutations and Combinations.In fact these two principles form the base of Permutations and Combinations.

Fundamental Principle of Multiplication

"If there are two jobs such that one of them can be completed in m ways, and another one in n ways then the two jobs in succession can be done in m X n ways."

Example :- In her class of 10 girls and 8 boys, the teacher has to select 1 girl AND 1 boy. In how many ways can she make her selection?Here the teacher has to choose the pair of a girl AND a boy

For selecting a boy she has 8 options/ways AND that for a girl 10 options/waysFor 1st boy ------- any one of the 10 girls ----------- 10 waysFor 2nd boy ------- any one of the 10 girls ----------- 10 waysFor 3rd boy ------- any one of the 10 girls ----------- 10 ways-------------------------For 8th boy ------- any one of the 10 girls ----------- 10 waysTotal number of ways 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 = 8b0 ways OR 10 X 8 = 80 ways.

Remark :-The above principle can be extended for any finite number of jobs.

Fundamental Principle of Addition

"If there are two jobs such that they can be performed independently in m and n ways respectively, then either of the two jobs can be performed in (m + n) ways."

Example :- In her class of 10 girls and 8 boys, the teacher has to select either a girl OR a boy. In how many ways can she make her selection?Here the teacher has to choose either a girl OR a boy (Only 1 student)

For selecting a boy she has 8 options/ways OR that for a girl 10 options/ways. The first of these can be performed in 8 ways and the second in 10 ways.Therefore, by fundamental principle of addition either of the two jobs can be performed in (8 + 10) ways. Hence the teacher can make the selection of a student in 18 ways.

Examples 1 :- There are 3 candidates for a classical, 5 for a mathematical, and 4 for a natural science scholarship.

I. In How many ways can these scholarships be awarded?

Clearly classical scholarship can be awarded to anyone of the 3 candidates, similarly mathematical and natural science scholarship can be awarded in 5 and 4 ways respectively. So,Number of ways of awarding three scholarshipsV= 3 X 5 X 4 = 60 -----------------------[ By Fundamental Principle of Multiplication]

II. In How many ways one of these scholarships be awarded?

Number of ways of awarding one of the three scholarships = 3 + 5 + 4 = 12------------------------[ By Fundamental Principle of Addition]

Example 2 :- A room has 6 doors. In how many ways can a person enter the room through one door and come out through a differentNumber of ways coming in the room = 6Number of ways going out of the room = 5 (He/She cannot go from the same door)

By Fundamental Principle of Multiplication--------> Coming in X Going out = 6 X 5 = 30.

Example 3 :- Five persons entered the lift cabin on the ground floor of an 8 floor house. Suppose each of them can leave the cabin independently at any floor beginning with the first. Find the total number of ways in which each of the five persons can leave the cabini) At any one of the 7 floorsii) At different floors.

Let the five persons be b,c,d,e,f

I) b can leave the cabin at any of the seven floors. So he has 7 optionsSimilarly each of c,d,e,f also has 7 options. Thus the total number of ways in which each of the five persons can leave the cabin at any of the seven floors is7 X 7 X 7 X 7 X7 =75

II) b can leave the cabin in 7 ways. c can leave the cabin in 6 ways, since he can not leave at where b left. In the same way d has 5, e has 4, and f has 3 way.Hence total number of ways = 7 X 6 X 5 X 4 X 3 = 2520

Example 4 :- In how many ways can 3 prizes be distributed among 4 boys, wheni) No boy gets more than one prize ?ii) A boy may get any number of prizes ?iii) No boy gets all the prizes ?

I) The first prize can be given away to any of the 4 boys, hence there are 4 ways to distribute first prize.The second prize can be given away to any of the remaining 3 boys because the boy who got the first prize cannot receive second prize.Similarly third prize can be given away to any of the remaining 2 boysHence total number of ways are 4 X 3 X 2 = 24Note :- This is same as Arrangement of 4 boys taken 3 at a time in a way 4P3 = 4!/1! = 4! = 24 (More on this under the Head of Permutations later)

II) First prize to any one of the 4 boys. Similarly second to any one of the 4 boys, and third as well to any one of the 4 boys = 4 X 4 X 4 = 43 = 64

III) Since any one of the 4 boys may get all the prizes. So, the number of ways in which a boy gets all the 3 prizes = 4So the number of ways in which a boy does not get all the prizes = 43 4 = 60

Permutations (Arrangement) nPr

Each of the arrangements which can be made by takingsomeorallof a number of things is called a permutation.

For example, if there are three objects namely x,y, and z,

then the permutations of these objects, taking two at a time, are xy, yx, yz, zy, xz, zx = Total 6 Permutations.

NOTE :- It should be noted that in permutations the order of arrangement is taken into account; When the order is changed, a different permutation is obtained.

Now Lets go to some theoretical part

Most of us know the popular formula of calculating number of permutations

It is nPr i.e. Number of all permutations allndistinct things takenrat a time (1 r n) = nPr =n!(nr)!

We will try to know how it is. We will try to prove it by a numerical example.

In how many different ways would you arrange 5 persons on 3 chairs

Here n= 5 and r= 3

Arranging 5 persons on 3 chairs is same as filling 3 places when we have 5 different things at our disposal.

The first place can be filled in 5 ways -------------(Remember Fundamental Principal of Multiplication)

Having filled it, there are 4 things left and anyone of these 4 things can be used to fill second place. So the second place can be filled in 4 ways.

Hence by fundamental principal of multiplication, the first two places can be filled in 5 X 4 ways

When the first two places are filled, there are 3 things left, so that the third place be filled in 3 ways

So the total number of arrangements will be 5 X 4 X 3.

As per the formula of permutation it will be 5P3 =5!(53)!=5!2!= 5 X 4 X 3

Remark :- Continuing in this manner we can say that number of permutations (or arrangements ) ofnthings taken all at a time will ben!e.g. In how many ways can 6 persons stand in a queue? Here n= r=6 so total number of permutations will be 6! = 720

Example 1 :- It is required to seat 4 Women and 5 Men in a row so that the women occupy the even places. How many such arrangements are possible?Total Places = 9Even Places = 4 i.e. 2nd, 4th, 6th, 8th4 women can be arranged on 4 even places in 4! Ways.5 men can be arranged on remaining 5 places in 5!

By Fundamental Principle of Multiplication, Total number of arrangements will be 4! X 5! = 24 X 120 = 2880

Example 2 :- Three men have 4 coats, 5 waist coats, and 6 caps. In how many ways can they wear them?

Number of ways in which 3 Men can wear 4 coats =4!1!= 4! = 24Number of ways in which 3 Men can wear 5 waist coats =5!2!= 5 X 4 X 3 = 60Number of ways in which 3 Men can wear 6 caps =6!3!= 6 X 5 X 4 = 120

By Fundamental Principle of Multiplication

Arrangement of 4 Coats AND Arrangement of 5 Waist Coats AND Arrangement of 6 Caps

Total number of ways = 24 X 60 X 120 = 172800

NOTE :- It is unlikely that GMAT will ask such an question in the real test, since it involves tedious calculations. However the idea behind putting this here is to show how the Fundamental Principle of Counting works.

Lets modify the above question

Three men have 4 coats, 5 waist coats, and 6 caps. In how many ways can they wear any one type of them?

Number of ways in which 3 Men can wear 4 coats =4!1!= 4! = 24Number of ways in which 3 Men can wear 5 waist coats =5!2!= 5 X 4 X 3 = 60Number of ways in which 3 Men can wear 6 caps =6!3!= 6 X 5 X 4 = 120

By Fundamental Principle of Addition

Arrangement of 4 Coats OR Arrangement of 5 Waist Coats OR Arrangement of 6 Caps

Total number of ways = 24 + 60 + 120 = 204

Example 3 :- How many four digit numbers are there with distinct digitsn = 10 i.e. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9and r = 4

Total Number of arrangements =10!(104)!

But these arrangements also include those numbers which have zero (0) at thousands place.Such numbers are not four digit numbers and hence need to be excluded.When 0 is fixed at thousands place, we have to arrange remaining 9 digits by taking 3 at a time in a way9!(93)!Hence total number of four digit numbers =10!(104)!9!(93)!= 5040 504 = 4536

Example 4 :- Find the sum of all the numbers that can be formed with the digits 2, 3, 4, and 5 taken all at a time.

Total Number of numbers = 4! = 24

To find the sum of these 24 numbers, we will find the sum of digits at Units, Tens, hundreds, and thousands places in all these numbers.

Keep 2 at units place. _ _ _ 2We can fill the remaining places in 3! i.e. 6 ways.So there are 6 numbers in which 2 can occur at units place.Similarly we can place all other digits at units places and will get 6 different numbers for each digit.So, total for the digits in the units place in all the numbers = 2 X 3! + 3 X 3! + 4 X 3! + 5 X 3!= (2 + 3 + 4 + 5) 3! = 84-----------Sum of Units Places (NOTE :- Above is an explanation for this direct formula)= (2 + 3 + 4 + 5) 3! X 10 = 840-----------Sum of Tens Places= (2 + 3 + 4 + 5) 3! X 100 = 8400-----------Sum of Hundreds Places= (2 + 3 + 4 + 5) 3! X 1000 = 84000-----------Sum of Thousands PlacesSum = 93324

By other Way

(2 + 3 + 4 + 5) 3! = 84 -------------> 84 (100 + 101 + 102 + 103) = 93324

Next Case :- What if repetition of digits would have been allowed?

Keeping 2 at units place. _ _ _ 2 we would get 4 X 4 X 4 = 64 numbers.(2 + 3 + 4 + 5) 64 = 896 -------> 896 (100 + 101 + 102 + 103) = 896 (1+10+100+1000) -------> 896 X 1111 = 995456

The similar example is discussed atwhat-is-the-sum-of-all-3-digit-positive-integers-that-can-be-78143.html#p1164866

Example 5 :- There are 6 periods in each working day of a school. In how many ways can one arrange 5 subjects such that each subject is allowed at least one period?

5 periods can be arranged in 6 periods in 6P5 ways. Now one period is left and it can be allotted to any one of the 5 subjects. So number of ways in which remaining one period can be arranged is 5.Total Number of arrangements = 6P5 X 5 = 3600

A Probability Question on the same logic is here.combinatorics-probability-problem-please-help-150364.html#p1209126

Permutations under Certain Conditions

Here we shall see permutations where either repetitions of items are allowed or distinction between some of the items is ignored or a particular item occurs in every arrangement etc.

First Some Theoretical ( boring?) Part

1) Permutations ofndifferent objects takenrat a time,when a particular object is to be always included in each arrangementisr(n-1Pr-1)

Objects = n

Places to be filled = r

Condition = A Specific Object (lets say x) should always be included in each arrangement.

We will place this x at first place. Now we are left with n-1 objects and r-1 places.

We can arrange the n-1 objects at r-1 places in n-1Pr-1 ways (NOTE Our X was at 1st place in these arrangements)

So each time we push x one place ahead by one position (i.e. 2nd, 3rd,4th,.rth), we will get n-1Pr-1 new ways of arranging n-1 at r-1 places.

How many places canxgo ?rplaces right!

So Total Permutations = n-1Pr-1 + n-1Pr-1 + n-1Pr-1 r times

=r(n-1Pr-1)

2) Permutations ofndifferent objects takenrat a time, when a particular object is never taken in each arrangement isn-1Pr

Since a particular object is never taken. So, we have to determine the number of ways in which r places can be filled with n-1 distinct objects.

Example 1 :- Make all arrangement of letters of the word PENCIL so thati) N is always next to Eii) N and E are always together.

I) Lets keep EN together and consider it one letter. Now we have 5 letters which can be arranged in a row in 5P5 = 5! = 120 ways.

II) Solve as above. Just keep in mind that now E and N can interchange their places in 2! Ways. So total arrangements = 5!2! = 240

Example 2 :- A Tea Party is arranged for 16 persons along two sides of a long table with 8 chairs on each side. 4 Persons wish to sit on one particular and 2 on the other side. How many arrangements are possible?

Let the Sides be A and B

Let 4 persons wish to sit at side A. They can sit in 8P4 ways.

Similarly 2 Persons can sit on other side in 8P2

Now we are left with 10 places and 10 persons. So they can be arranged in 10! Ways.

Total Number of Arrangements = 8P4 X 8P2 X 10!

Permutations of Objects not all distinctn!p!q!

Many of us may be familiar and adept in solving problems pertaining to this concept

For example if I ask, How many words (with or without meaning) can be formed using all digits of the wordINDIA

Almost everybody will say it is5!2!

Why we need to do this division?

We will first look the Theorem underlying above

Theorem :- The number of mutually distinguishable permutations ofnthings, taken all at a time, of whichpare alike of one kind,qare alike of second such that p + q = n, isn!p!q!

Lets saynis such that n=p+q wherepis set ofpalike things andqis set ofqalike things

Letsxis the total permutations ofn

Let n=5, p=3, q=2 -----------------lets consider the integers 3,3,3,2,2 where n=33322, p=333, q=22

Now Replacepalike things bypdistinct things--------------- Replace integers 3,3,3 by integers 4,5,6

These replaced integers 4,5,6 could have been permuted themselves inp!or 3! or 6 ways.

Similarly Replaceqalike things byqdistinct things--------------- Replace integers 2,2 by integers 7,8

These replaced integers 7,8 could have been permuted themselves inq!or 2! or 2 ways.

Now we knowxis the total number of permutations ofn.We also know number of permutations ofpis 1 and that ofqis also 1. (Number of permutations of 3, 3, 3 = 1 or of 2, 2 = 1)

If all the terms in set P and Q were different the total number of permutations ofnwould have beenx X p! X q! = n! --------> x X 3! X 2! = 5! -----------> x X 6 X 2 = 120 ---------x X 12 = 120

So x =n!p!q!---------- x =12012------> x=10

REMARK :- The total number of permutations ofnthings, of whichpare alike of one kind,qare alike of second kindand remaining all are distinct, isn!p!q!

Example 1 :- How many words would you form with the letters of the word MISSISSIPPITotal =11, S = 4 times, I = 4 times, P = 2 timesSo11!4!4!2!= 34650

Example 2 :- APPLETotal = 5, P = 2 times, A = 1 time, L = 1 time, E = 1 time5!2!1!1!1!=5!2!= 60

Circular Permutations (n-1)!

So far we have discussed linear permutations. The basic difference between linear permutations and circulations is that every linear arrangement has a beginning and an end, but there is nothing like beginning or end in a circular permutation. Thus, in circular permutations, we consider one object as fixed and the remaining objects are arranged as in case of linear arrangement.

Thus Total number of circular arrangement of n distinct objects is (n-1)!

Note :- There are certain arrangements in which clockwise and anticlockwise arrangements are not distinct. e.g. arrangements of beads in a neckless, arrangement of flowers in a garland etc. In such cases number of circular permutations of n distinct objects is [(n-1)!]

Example :-i) Arrange 5 persons around a circular table

(5-1)! = 4! = 24

ii) In how many of these arrangements will two particular persons be next to each other?

Consider two particular persons as one person. We have 4 persons in all. These 4 persons can be seated around a circular table in (4-1)! = 3! Ways. But two Particular persons can be arranged between themselves in 2! WaysSo Total number or arrangements = 3! X 2! = 12

Example 2 :- Solve this question in light of above explanation. OA and OE can be found in spoiler.If 20 persons were invited for a party, in how many ways will two particular persons be seated on either side of the host?A) 19! X 2B) 18! X 3!C) 18! X 2D) 20!/2!E) 21!/2!

[Reveal]Spoiler:

Example 3 :- In how many ways can a party of 4 men and 4 women be seated at a circular table so that no two women are adjacent?

The 4 men can be seated at the circular table such that there is a vacant seat between every pair of men in (4-1)! =3! Ways. Now 4 vacant seats can be occupied by 4 women in 4! Ways. Hence the required number of seating arrangements = 3! X 4! = 144

Combinations (Selection) nCr

So far we have discussed Permutations (arrangements) of a certain number or objects by taking some of them or all at a time.

Most of the times, we are only interested in selection of objects and not their arrangements. In other words, we do not want to specify the ordering of selected objects.For example, a company wants to select 3 persons out of 10 applicants, a student wants to choose three books from his library at a time etc.

Suppose there are three objects namely x,y, and z.

We are asked to calculate the permutations (arrangements) of these objects taking 2 at a timexy, yx, yz, zy, xz, zx = Total 6 Permutations.

Now we are asked to calculate the combinations (selections) of these objects taking 2 at a timexy, yz, xz = Total 3 Combinations.

Note the important difference here. In later case we did not differ between xy and yx, as we did in the first case. This is the only difference between Permutations and Combinations.

In Combinations we find different ways of choosingrobjects fromngiven objects

While in Permutations we find different ways of choosing r objects from n given objects and ways of arranging these r objects.

The formula of permutations(nPr) itself says first selection (nCr) and then arrangement (r!)

nPr = nCr X r!

n!(nr)!=n!r!(nr)!X r! ----------->n!(nr)!=n!(nr)!

Difference between A Permutation and Combination.

1. In a combination, the ordering of the selected objects is immaterial, whereas in a permutation, the ordering is essential. For examplea,bandb,aare same as combinations but different as permutations.

2. Each Combination Correspond to many permutations. For example, the six permutationsABC, ACB, BCA, BAC, CBA,andCABcorrespond to the same combinationABC

3. Generally we use the wordarrangements for permutationsand the wordselections for combinations.

Example 1 :- From a class of 32 students, 4 are to be chosen for a competition. In how many ways can this be done?32C4 =32!28!4!

Example 2 :- 3 Gentlemen and 3 Ladies are candidates for 2 vacancies. A voter has to vote for 2 candidates. In how many ways can one cast her/his vote.

In all there are 6 candidates and a voter has to vote for any 2 of them. So he can select 2 candidates from 6 candidates in 6C2 ways =6!4!2!=(6X5)2= 5 X 3 = 15

Example 3 :- In how many ways can a cricket eleven be chosen out of a batch of 15 players if

i) There is no restriction on the selection = 15C11

ii) A Particular Player is always chosen = 14C10

iii) A Particular Player is never chosen = 14C11

Example 4 :- Out of 2 Women and 5 Men, a committee of 3 is to be formed. In how many ways can it be formed if at least one woman is to be included?

A Committee can be formed in the following ways

1 woman AND 2 men OR 2 women AND 1 man

2C1 X 5C2 + 2C2 X 5C1 = 20 + 5 = 25

Example 5 :-(Pay special attention here, because things to be arranged are identical)In how many ways can 7 plus(+) signs and 5 minus (-) signs be arranged in a row so that no two minus signs are together?

The Plus signs can be arranged in only one way, because all are identical.

| |+| |+| |+| |+| |+| |+| |+| |

There are 8 boxes available for placing 5 minus signs. So 5 boxes can be chosen out of 8 boxes in 8C5 ways.

However since 5 minus signs are identical, they can be placed in 5 chosen boxes in only one way

So Total number of possible arrangements = 1 X 8C5 X 1 = 56

Other Important Concepts in Permutations and Combinations

1)2n= Selection of any number of things out ofndistinct things.

Let there bendistinct things and you have told to choose any number things out ofnthings

each thing you can either select or deselect.

So you get two options for dealing with each thing.

There arenthings and you have 2 options for each thing.

In all you will have 2 X 2 X 2 X 2 X 2 Xn times options -------By Fundamental Principle of Multiplication

So there will be2noptions with you for selection

Example :- Number of ways Shreya can or cannot eat sweets at a party out of 10 distinct sweets available at party =210

Corollary =2n 1 = When zero selections are not allowed. i.e. If Shreya has asked to select at least one sweet.

Note :- This is especially important in SET concept.

For example let the Set A = (1,2,3)

Question :- How many subsets does the Set A have.

Solution :- We know a subset can be formed by taking any number of elements (or even zero) from its superset.2nstands for selection of any number of things out of given things.

So the number of subsets of Set A will be23= 8 (Here n=3 because Set A have 3 distinct elements)(0) (1) (2) (3) (1,2) (2,3) (1,3) (1,2,3)

2) n+1 = Selection of any number of things out of n identical things

Selection of any number of balls from 3 red balls.

Zero ball selected 1 way

One ball selected 1 way

Two balls selected 1 way

Three balls selected 1 way

Total (3 +1 ) = 4 ways.

3)n(n3)2= Number of diagonals of n sided polygon

4) nC2 - rC2 + 1 = Number of straight lines formed bynpoints of whichrare collinear

5) nC3 - rC3 = Number of triangles formed bynpoints of whichrare collinear

6) Withmparallel lines intersected bynparallel linesmn(m1)(n1)4parallelograms can be formed.Note :- This formula is derived from mC2 X nC2

7) Sum of the Numbers formed by using n digits taking all at a time is (n-1)! (1111..n times) (Sum of Digits)

8) Ifnis even, then nCr will be maximum for r =n2