Hi all,
Permutation& Combination and Probability:
Permutation and Combination are not that important for the purpose of exam Because Question are rarely asked from This Topic but We have to learn them anyway because Question of probability can’t be solved without learning permutation and combination. So will give you all a little hint about what is permutation and what is combination and then we will move on to Probability.
But Before That Just Look at A very Important Concept Without Which You can’t solve a single Question of permutation/combination or probability.
And that Factorial Notation.
It’s represented by (!) and it is read as Factorial.
So if i write 5! it will be read as Five Factorial.
And what it means ? It means to simply multiply all the numbers in decreasing order till 1.
Like if i write 6! it means 6*5*4*3*2*1 = 720
Or 7! = 7*6*5*4*3*2*1 = 5040
For Fast Calculation You all must learn the value of factorial till 10.
Just Learn these values
1! = 1
2! = 2
3!= 6
4! = 24
5! = 120
6! = 720
7! = 5040
8! = 40320
9! = 362880
10! = 3628800
Well Before I Start Explaining Permutation and combination one thing i want to tell and that is It’s the easiest topic that you will find in maths. Most people are unable to understand it and that’s why people think it’s complex and all type of misconceptions but trust me it’s the easiest topic in the whole mathematics and It’s not actually even maths, It’s less about Calculation and more about Logical Thinking. Well We all can’t Calculate Fast but we all can think fast.
So what is permutation?
In Simple words it’s arrangement or No. of ways things can be arranged.
Suppose there are 3 words ABC and if it’s asked How many ways these three can be arranged then all yu or What are the no. of permutations Possible. Then all you have to do is Arrange this things in as many ways it’s Possible.
Let’s try to arrange them now. SO There is ABC, ACB, BAC, BCA, CAB, CBA Are there any more ways these can be arranged ? try it ? No These are the all possible arrangements. So The answer to the above Question will be 6. That is ABC can be arranges in different ways.
Now there were only 3 alphabets What if there were more like You have to Arrange ABCDEFGHI. Now for 3 alphabets it was easy you easily arranged them But Arranging these 9 letters will take you days and even then you will not be able to get a certain answer.
So what we should do here. No need to worry our mathematicians were genius they created a very simple formula for that.
And Formula is like this.
N Different things can be arranged in N! ways.
So in above Question there were 9 alphabets so the no. of possibele arrangements will be 9! = 362880.
So that was out basic concept Now let’s move on to another basic concept.
So in the above questions It was Asked in how many ways ABCDEFGHI Can be arranged. In this question they were asking the possible arrangements of all the 9 Alphabets, They can also Ask In how many ways 4 alphabets from above 9 alphabets can be arranged.
In such type of Questions there is another formula Which is very very very important because it will be used in almost every question.
So the formula is Out of n things r things can be arranged in nPr ways. and
nPr = n!/(n-r)!
So in the above Question it is asked that in how many ways 4 alphabets from the total 9 alphabets can be arranged.
So apply the formula nPr = 9P4 = 9!/(9-4)! = 9*8*7*6*5*4*3*2*1/5*4*3*2*1 =9*8*7*6 = 3024.
Now there is a trick to easily calculate nPr by which you won’t have to do any division work.
Like if it say 9P3 then you just have to multiply Starting from 9 in decreasing order till the next 2 digit i.e 9P3 = 9*8*7. Why we multiply till 7 only ? that is because the value of r is 3 and total multiplication should contain the value of r.
Another example if it 7P2 then you will just do 7*6[ 2 number because r = 2 ok]
if it’s 7P4 then the answer will be 7*6*5*3[ 4 no. because value of r=4]
So If it’s 10P5 then the value will be 10*9*8*7*6 [ 5 digit because value of r = 5]
I think you understand my point now. Now move on to the cases.
Actually there are infinite cases in Permutation and Combination 100’s of different type of question can be formed So i will only discuss the cases that are important for the exam, And if you have any problem in any other case then you can ask me personally.
Case – 1 Simple Arrangement Case well all words are unique.
By UNIQUE i mean all alphabets are different
In how many ways the letters of the word ROCKET can be arrnged.
very Simple just count the no. of words in ROCKET that will be 6
So number of arrangements will be n! that will be 6!
CASE – 2 Arrangement When All the words are not UNIQUE.
That means some words are repeated.
Like No. of possible arrangements of word TITANIC
Now In this case you Just have to find the total possible ways first without even thinking about Repeated words and then after that You will divide that with the numbers of times a Word is repeated.
So in the above Question Total alphabets = T = 2, I = 2. A= 1 C =1 N = 1Total 7 So Permutations will be 7! and Now you will divide It by No. of times A word is repeated SO T is repeated 2 times and I is repeated two times So divde 7! by these 2. So final Answer will be 7!/(2!*2!)
Let’s See another Example. In how many ways the letters of the word RUNNING Can be arranged.
So total no. of alphabets in the above Words = 7
No. of words that are repeated = N = 3 times repeated.
So the solution will be Total permutation divided by no. of times a word is repeated and that will be 7!/3! that will be your answer.
Case 3 – Arrangement Some Words are always together and Some Words and Never together.
No of possible arrangements of the words LAYERING When Vowels are always together.
In this case what we do Is we consider the no. of Vowels as 1 single alphabet That [AEI] is a one single alphabet In that way they will always be together and the rest words are LYRNG.
So the total no. of alphabets will be 6 ? Why 5 Alphabets are LYRNG and [AEI] is One alhpabet remember so The total alphabet will be 6
And no. of possible arrangements will be 6!
But but the question is not complete yet [AEI] Though considered as 1 alphabet but stil the words AEI can change places within itself Like AEI it also can be AIE or EIA. So there are 3 words so no. of total arrangements that they can do within itself will be 3!
So our final answer will be 6!*3![ that is because 6! is the no. of possible ways when AEI are together and And multiplied by 3! because AEI can change places within themselves in 3! possible Ways]
If it was asked that VOWELS in LAYERING are never together that what we will do ?
This Question can’t be solved directly.
In order to solve this We will have to FIND the total no. of arrangements of the word LAYERING and then Subtract the no. of arrangemnts in which AEI are Always together.
So no. of possible arrangements of LAYERING will be 8!
And We already Solved that when AEI are always together the no. of possible ways are 6!*3!
So no. of possible ways when AEI are never together will be 8! – 6!*3!
Now i told you that there are many more cases but that are really not important I am explaining these cases because they are important and help ypu while solving Probability.
Now We should move on to the next Topic That Is Combination. Now you know that Permutation means Arrangement or no. of possible ways A thing can be arranged.
What is the meaning of Combination.
Combination is a simple act of Choosing or Selection.
Like When it is asked What are no. of possible ways Word TITAN can be arrange You have to find The Permutation.
But if it is asked what are no. of possible ways You can Select 2 alphabet from the word TITAN, It means you have to find Combination.
The act of selection or Choosing is called COMBINATION.
Now you all must know what is nPr so it’s time to move towards nCr
Like nPr = n!/(n-r)!
nCr is somewhat simillar but that is just an extra r! in the denominator
So nCr = n!/[(n-r)!*r!]
nCr means r things has to be selected out of n things.
Like in the above Question No. of possible ways 2 alphabets can be selected from the word TITAN
So total no. of alphabets n = 5
no. of alphabets which we have to select r = 2
So the answer will be 5C2 = 5!/(5-2)!*2! = 5!/3!*2! = 5*4/2*1 = 10
Now i told you have to calculating nPr in a simple way Just like that we can also calculate nCr in a simple way All you have to do is Follow the method of nPr and In division you have to also multiply in increasing order from 1
Like 6C3 = 6*5*4/1*2*3
And 9C2 = 9*8/1*2
and 10C4 = 10*9*8*7/1*2*3*4
7C5 = 7*6*5*4*3/1*2*3*4*5
This much knowledge of combination is enough for solving the Questions of Probability.
So without wasting Time just move on to our main Topic ie Probability.
Probability
So what is Probability ?
Probability is Just the chances have happening of an event. Like what are the chances that You will Become a PO or An Income Tax Inspector or a Clerk. What are the chances that you will find the love of you life (That chance of that is very rare)
These all chances are just the game of Probability. Our Life is Also The sum of all these chances, the chances we take Like What are the chances that you will study after 12 instead of gossiping on whatsapp.
So how de we find the probability of happening of an event. In mathematical terms probability = Number of favourable Outcomes/ Total outcomes
No. of favourable outcomes means the outcomes which we want.
Total outcomes Means the total possible outcomes (That’s the reason we studied Permutation and Combination so that we can find total outcomes]
Let me give you a very realistic example. What is the probability that You will Become a PO in SBI ?
So We have to find the favourable outcomes here That will be the No. of Posts in SBI[ because if you get any of the post in the total post you will be a PO]
So total no. of Posts In SBI this time is 2000
And what are the total outcomes or What are the total no. of Applicants = 20,00,000
So what is the probability that You will be 1 of them Simple Probability of You getting selected = favourable Outcomes/ total outcomes = 2,000/20,00,000 = 1/1000
That is Your Chances. Or in other words 1 in a thousand Aspirant can become a PO in SBI.
So i think Now you have the basic Idea what is PROBABILITY.
So now Lets Move On to Questions.
But before that.VERY VERY VERY VERY VERY IMPORTANT
AND = Multiplication(*)
OR = Addition (+)
If anywhere and I mean Anywhere you see a question which say what is the probabilty of getting X or Y, It simply means that you have to find probabilty of X and Probability of Y and ADD them, The word OR means Addition Always Keep in Mind that.
And if It is asked what is the prbability of getting X and Y, It simply means that you have to find the probability of X and Y and Multiply them, The word AND means Multiplication Always remember that.
At least = Minimum We require [ Or kam se kam Kitna hone chahiye Usase jyada bhi ho sakta hai but usase kam nahi hona chahiye]
Example If we want at least 2that measn Minimum we need 2 We can have 3 or 4 or 5 It doesn’t matter but Should not be less than 2.
At Most = Maximum We Require[ Jyada se Jyada Kitna ho sakta hai, Usase Kam ho sakta hai farak nahi padta but usase jyada ahi hona chahiye]
Example if we want AT MOST 2 That means we can have 2 we can have 1 and we can have 0 also any less value it doesn’t But we can can’t have anything greater than 2.
These cases will be more clear to you when we will solve some Questions.
Questions related to Balls.
Case 1: Normal Case
There are Total 5Red, 3Blue and 2 Green bals In a Bag, Two balls are taken out at random What is the probability that
i)-2 Balls will be Green.
ii- 2balls will be RED
iii) – 2 balls will be BLUE.
i) What is the probability that 2 balls are taken out at random from a bag and both balls are Green.
So Calculate First the favourable Outcomes. That is how many ways 2 balls can be taken out from a bag which have 2Green balls = 2C2 = 1
Now calculate Total Outcomes. That is how many ways 2 balls can be taken out from the bag containing total 10 balls [ 5red + 3Blue + 2 Green = Total 10] = 10C2 = 10*9/1*2 = 45
So probability = favourable outcomes/total outcomes = 1/45
ii) What is the probability that 2 balls are taken out from bag and both are RED
So Calculate First the favourable Outcomes. That is how many ways 2 balls can be taken out from a bag which have 5Red balls = 5C2 = 5*4/1*2 = 10
Now calculate Total Outcomes. That is how many ways 2 balls can be taken out from the bag containing total 10 balls [ 5red + 3Blue + 2 Green = Total 10] = 10C2 = 10*9/1*2 = 45
So probability = favourable outcomes/total outcomes = 10/45 = 2/9
iii) What is the probability that 2 balls are taken out from bag and both are Blue.
So Calculate First the favourable Outcomes. That is how many ways 2 balls can be taken out from a bag which have 3BLUE balls = 3C2 = 3*2/1*2 = 3
Now calculate Total Outcomes. That is how many ways 2 balls can be taken out from the bag containing total 10 balls [ 5red + 3Blue + 2 Green = Total 10] = 10C2 = 10*9/1*2 = 45
So probability = favourable outcomes/total outcomes = 3/45 = 1/15.
CASE 2 – AND Case.
There are Total 5Red, 3Blue and 2 Green bals In a Bag, Three balls are taken out at random What is the probability that
i) 2 balls are Red and 1 ball is Green
ii) 2balls are Blue and 1 ball is Green
i) What is the probabilty that 3 balls are taken out and out of those 3 balls 2 balls are red and 1 is green.
In this questions there are 2 events i.e getting 2 red ball and getting 1 green ball
So first we have to calulate the seperate probabilities first.
So no. of ways 2 Red balls can be selected out of total 5 balls = 5C2 = 5*4/1*2 = 10
So no. of ways 1 Green ball can be selected out of total 2 balls = 2C1 = 2
So Favourable Outcomes i.e No. of ways 2 Red balls AND 1 Green Ball can Be Selected = 10*2 = 20[ Whenver you see and Just Multiply it ]
And Total No. of Outcomes i.e Selecting 3 balls out of total 10 balls = 10C3 = 10*9*8/1*2*3 = 120
So probability of getting 2 Red and 1 green Balls = favourable outcomes/total outcomes = 20/120 = 1/6
ii) What is the probabilty that 3 balls are taken out and out of those 3 balls 2 balls are BLUE and 1 is GREEN
So no. of ways 2 BLUE balls can be selected out of total 3 balls = 3C2 = 3*2/1*2 = 3
So no. of ways 1 Green ball can be selected out of total 2 balls = 2C1 = 2
So Favourable Outcomes i.e No. of ways 2 BLUE balls AND 1 Green Ball can Be Selected = 3*2 = 6[ Whenver you see and Just Multiply it ]
And Total No. of Outcomes i.e Selecting 3 balls out of total 10 balls = 10C3 = 10*9*8/1*2*3 = 120
So probability of getting 2 BLUE and 1 green Balls =favourable outcomes/total outcomes = 6/120 = 1/20
Case 3: OR CASE
There are Total 5Red, 3Blue and 2 Green bals In a Bag, 2 balls are taken out at random What is the probability that 2 balls are Red or 2 balls are Blue
In this questions there are 2 events i.e getting 2 red ball or getting 2 Blue balls
So first we have to calulate the seperate probabilities first.
So no. of ways 2 Red balls can be selected out of total 5 balls = 5C2 = 5*4/1*2 = 10
So no. of ways 2 BLUE balls can be selected out of total 3 balls = 3C2 = 3*2/1*2 = 3
So Favourable Outcomes i.e No. of ways 2 RED balls OR 2 BLUE Balls can Be Selected = 10 + 3 = 13[ Whenver you see OR Just ADD it ]
Now calculate Total Outcomes. That is how many ways 2 balls can be taken out from the bag containing total 10 balls [ 5red + 3Blue + 2 Green = Total 10] = 10C2 = 10*9/1*2 = 45
So the probability of getting 2 Red ball or 2 Blue balls = favourable outcomes/total outcomes = 13/45
CASE 4 – AT LEAST CASE
There are Total 5Red, 3Blue and 2 Green bals In a Bag, Three balls are taken out at random What is the probability that At least 2 Balls are RED.
Now What i Told you In At Least Case You have to select at least 2 means You can have all 3 balls red But at least 2 balls should be RED means we will have to find the probability of getting 2 red balls OR 3 red balls.
So there are 2 cases here 1st case is when we get 2 red balls and 1 ball can be of any other colour
and 2nd case is when we get all 3 balls as red.
1st case
So no. of ways 2 Red balls can be selected out of total 5 balls = 5C2 = 5*4/1*2 = 10
And No. ways 1 ball can be selected out of rest 5 balls = 5C1 = 5
Our Favourable outcomes I.e getting 2 red balls and 1 ball of any colour = 5*10 = 50 [ And case so multiply ]
And Total No. of Outcomes i.e Selecting 3 balls out of total 10 balls = 10C3 = 10*9*8/1*2*3 = 120
So Probability of getting 2 red balls and 1 ball of any other colour = favourable outcomes/total outcomes = 50/120 = 5/12
2nd case When we get all 3 balls as red.
So no of ways 3 red balls can be selected out of total 5 red balls i.e also our favourable outcome = 5C3 = 5*4*3/1*2*3 = 10
And Total No. of Outcomes i.e Selecting 3 balls out of total 10 balls = 10C3 = 10*9*8/1*2*3 = 120
So probability of getting 3 red balls =favourable outcomes/total outcomes = 10/120 = 1/12
Now Either Case One will happen OR Case 2 will happen. that means either we will get 2red balls and 1 other ball or we will get all 3 red balls So As i already explained that In OR case Probabilities gets added so we will just add the probability To get the final probability.
So when 3 balls are taken out at random the probability that at least 2 balls are green = 1/12 + 5/12 = 6/12 = 1/2
Case 5 At MOST CASE
There are Total 5Red, 3Blue and 2 Green bals In a Bag, Three balls are taken out at random What is the probability that At Most 2 Ball is RED.
So as i told you all in case of AT most We can have any number less than But not greater than That means We can Have 2 Red balls out of 3 balls and We also can 1 red ball out of 3 balls and we can also have 0 red balls but we can’t have More than 2 Red ball. That means all 3 balls can’t be RED.
So we will solve same like the last case.
No. Of ways 2 red balls and 1 others balls can be selected = 5C2*5C1 = 10*5 = 50
No. Of ways 1 red balls and 2 others balls can be selected = 5C1*5C2 = 5*10 = 50
Now Of ways Balls are selected that there are NO red balls That means All three balls are of Other Colours = 5C3 = 10
Total No. Of Outcomes = 10C3 = 10*9*8/1*2*3 = 120
So Probability will be (50+10 +50)/120 = 110/120 = 11/12
Source: bankersadda.com
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