Hello everyone, welcome back to our new video.
In today's video, we are going to do the work of
permutation and combination.
In the last video, we did the basic definition of
permutation and combination.
We learned the formulas.
And we had to find the difference between them.
We did the multiplication principle and addition principle.
I hope you know the difference between the two.
Where we will use multiplication and where we will use
addition.
In today's video, we will solve some questions and problems
on these.
We will see how to solve these problems.
So, let's move on to the questions.
So, let's see what is the first question in today's class.
See, the question is
In how many ways can we select a team of 4 students from a
given choice of 15?
See the given here.
How many students do we have in total?
We have 15 students in total.
What do you have to select from these? You have to select
from these.
We have to select 4 students.
You have to select only, not arrange.
There will be no difference in the arrangement.
You can select the first student or the fourth student.
It does not matter.
So, what will we use in this?
Combination.
And mostly, in the questions that you get,
Mostly, we use combination.
Permutation, as such, is asked less in the questions rather
than combination.
I am not saying that permutation will not come.
Permutation comes, but it is mostly used in combination.
So, selecting 4 students among 15 students.
What do we have to do?
We have to select, not arrange.
So, what will be our formula?
The value of n is 15.
The value of r is 4.
So, 15c4.
This will be our number of ways.
You can easily solve this.
15 factorial divided by, I am opening the formula.
15 minus 4 factorial into 4 factorial.
So, this will be 15 factorial upon 15 minus 4.
That is, 11 factorial into 4 factorial.
Now, see.
Above is 15 and below is 11 factorial.
So, this will intersect at some point.
Because, 15 will come up to 11 factorial.
So, how can we write 15?
If you start writing 15 backward.
15 factorial.
15 into 14 into 13 into 12 into 11 into 10 into 9 into 7.
9 into so on 1.
So, I can write this as 11 factorial.
Because, after this, the whole 11 factorial will open.
So, I forgot to write this property in properties.
I am sorry.
You can write this.
How can we write n factorial? n into n minus 1 factorial.
n into n minus 1 factorial.
We can simplify n factorial.
Or, we can do it backward.
Next, solve this.
11 factorial cancel.
Then, you solve 4 factorial.
So, you can write 4 factorial as 24.
This was a direct formula.
Or, 4 into 2 into 3.
So, 4 factorial will be 24.
So, let us solve from 24.
12 ones are 12 twos are.
Then, 2 will also cancel out from 14.
So, 2 ones are and 2 sevens are.
Now, the multiplication has become easy.
That is, you have to do 15 into 7 into 13.
I hope, you will get the answer.
This is a calculation.
I am leaving the calculation.
So, you can calculate this and write the answer.
Let us move to the next question.
This was basic.
How will the question come?
Let us move ahead.
Read the question.
In how many ways can a group of 5 members be formed by
selecting 3 boys out of 6 boys and 2 girls out of 5 girls?
You have to form a group of 5 members.
Total should be 5 members.
You need 5 members.
You have to select.
How? You have to select 3 boys.
There will be 3 boys in that group.
What will happen in this group?
There will be 3 boys.
And, what will be the rest?
There will be 2 girls.
Now, you have to select 3 boys out of 6 boys.
So, you have 6 boys.
You have to select 4 boys out of 6 boys.
What will be the result?
You have to select.
You don't have to arrange.
So, you will get 6C4.
You are choosing 4 boys out of 6 boys.
You have to select 3 boys.
So, you will get 6C3.
These are the number of ways of selecting 3 boys out of 6
boys.
So, you have to select boys.
We will do the same for girls.
So, how many girls do we have in total?
We have 5 girls in total.
And, how many girls do we have to select?
Select 2 girls.
So, we have 5 girls in total.
How many girls do we have to select out of 5 girls? 2 girls.
So, 5C2.
These are the choices to select 2 girls.
Now, what will we use?
We have 2 girls.
Now, we need 5 boys.
We need 7 boys.
And, we also had AND above.
So, what will be the result?
Multiply these 2 girls.
The answer will be 6C3 into 5C2.
You will get the answer to this question.
So, solve it.
You will get 6 factorial.
6 minus 3 factorial.
That is 3 factorial.
And, 3 factorial.
Multiply it by 5 factorial.
Divide it.
5 minus 2 factorial.
That is 3 factorial.
And, 2 factorial.
So, cancel out.
Take 6 back.
You can write 6 as 5 into 4 into 3 factorial.
So, cancel 3 factorial.
You can write 5 as 5 into 4 into 3 factorial.
So, cancel 3 factorial.
I am erasing the cancels.
3 factorial.
What is left? 3 factorial.
Which is 3 into 2 into 1.
6.
So, 3 factorial.
6 will cancel out.
2 factorial.
2 into 1.
So, 2.
And, it will be 2 from 4.
What do you have?
You have 5 into 4 which is 20.
Into 5 into 2 which is 10.
The answer is 200.
You have total 200 ways.
With which you can do this selection.
Which selection? 3 boys out of 6 boys.
And, 2 girls out of 5 girls.
It was simple.
If you understand their definition and formulas.
Then, you will understand the principle.
If you know what to do, then it is simple.
You are studying like a word problem.
You are solving.
Next.
See the question.
Okay.
This is a different question.
That is.
The number of positive integers which can be formed by
using any number of digits from 0, 1, 2, 3, 4, 5.
Without repetition.
This is to be taken care of.
Without repetition.
So, whenever you will get this combination, permutation
question.
So, in that we will have 2, 3 categories different.
Which are those 2, 3 categories?
Like, if you are selecting r objects from n.
So, you write ncr.
Now, these r objects.
Like, you have n objects.
You can have distinct objects.
You can have similar objects.
Means, same objects can be there.
Or, different objects can be there.
Let's suppose, you have n distinct objects.
Different objects.
You have n distinct objects.
You can have similar objects. Let us see how to choose R
objects from the list of alphabets.
Suppose you have 5 alphabets A, B, C, D, E.
You have to make a 3 letter word with the help of these 5
letters.
You have to find out how many ways you can fill up these
letters.
You have these 5 distinct objects A, B, C, D, E.
Now you get 2 cases.
You need 3 letters.
These 3 letters will be selected from these 5 letters.
Now you get 2 cases.
Either you can repeat these 3 letters.
Suppose you get 2 out of 3.
Suppose you get A here, A here, D here.
You get A, A.
This means that repetition is allowed.
You can select the object you are selecting later.
That is repetition is allowed.
So that was the case of repetition.
Whose repetition?
Next case is that repetition is not allowed.
If you fill up A here, you cannot fill up A anywhere else.
Later you will fill up some other number or letter.
So these were the two possibilities.
The third possibility is that you do not have these
distinct objects.
N is not distinct.
It is a similar kind of word.
Suppose you have 4 meats.
The words used in this meat are E.
E is repeated twice.
But you have to select from these 4.
E is repeated.
So what will you do?
You will fill up some other formula.
So these were the 3 distinct cases.
Let us note down these 3 distinct cases.
So what are these 3 distinct cases based on? First of all,
it is based on N.
Either N will be distinct.
Either N objects will be distinct.
Or they can be the same.
Should I write it like this?
Let me write it here.
Either N will be the same.
Or few are repeating.
For example, E was repeated in the meat.
M and T were distinct.
So these few objects are repeating.
Now in the case of distinct, you will have two sub-cases.
The first sub-case is that you will have repetition.
Repetition is allowed.
If repetition is allowed,
You have to select R objects.
R can be anything.
So you can put N ways here as well.
You can put anything from N.
So this is the position to fill up N ways.
You can put anything in the second position as well.
Suppose you have chosen A here.
So is it necessary that A cannot come here?
Obviously, if repetition is allowed,
A can come here as well.
It can come here as well.
It can be used again somewhere.
Or R bar can be used as well.
So here also you will have N ways to choose.
And here also N ways.
In this way, you will get total R times N.
How many times will you get? R times.
Because you have to fill up R position.
So whenever your repetition is allowed,
Your formula will be N to the power R.
This will be your direct formula.
I am telling you that you should remember the formula.
If you will follow the basic concept, There is no necessity
to remember this formula.
You can fill it up here.
You will get the answer.
Second is when repetition is not allowed.
When repetition is not allowed,
Then it cannot go to the place of these others.
Suppose you can fill this position with N ways.
There are total 10 ways.
There are 10 objects.
You can put any of these 10 objects.
So 10 objects become 10 ways.
Now here one object is used.
How many objects are left?
Only 9.
So here you can put only 9 objects.
So one of N will be less.
N minus 1.
Then here one more will be less.
So N minus 2 and so on.
This will continue till R.
Wherever R will go,
This will be your direct formula.
You can select as many terms as you want.
So this is when repetition and non-repetition is distinct.
Now suppose you have N objects.
Some of the objects are repeating P times.
Like E was repeating 2 times.
Similarly you can make any other word.
For example, you can make Mathematics.
Mathematics.
See how many objects are there in total? 1, 2, 3, 4, 5, 6,
7, 8, 9, 10, 11.
In total you have N 11.
Now which one is repeating?
One is M repeating 1 time, 2 times.
So M is repeating 2 times.
A is repeating 2 times.
T is repeating 2 times.
Is anyone else repeating?
No one else is repeating.
Rest all are repeating 1 time only.
So whichever number of times they are repeating,
We will call it M.
P is repeating 2 times.
So P's value is 2.
How many times is T repeating?
Let's suppose Q is repeating 2 times.
So Q's value is also 2.
Similarly, we will write all the repetitions as P, Q, R.
So the formula for this will be N factorial total divided by
Whatever number of times something is repeating, You can
write its factorial as
Divide by.
Whatever number of times something is repeating,
You can write its factorial as
Divide by.
For example, M is 2 factorial, so 2 factorial.
2 factorial will be repeated.
For example,
Like you have this mathematics,
You have to make words from this, Using all the letters of
11.
So the number of ways you can make letters,
Total number of words formed
Using the letters from this word mathematics.
What will happen?
N factorial is 11 factorial.
Divide by.
Divide by 2 factorial.
If it is repeating,
Then divide by 2 factorial.
Divide by 2 factorial.
I hope all three cases are clear to you.
So let's move on to the next question.
Because of which we came here.
So here is the question.
The number of positive integers,
You have to make positive integers,
Using any number of digits,
From 0, 1, 2, 3, 4 and 5.
Without repetition.
If you have used one digit earlier,
Then you can't use it again.
Like you can't make this number.
You can't make 11.
You can't make 222.
Because 222 is repeating.
These are the integers.
But they are repeating.
So we won't take this type of integers.
We will see what will be made other than these.
So how many total digits do you have? 6, 1, 0, 1, 2, 3, 4,
5, 6.
You have to make positive integers.
That's it.
Now he didn't say,
How many digits of positive integers you have to make.
So when he didn't tell you,
So all the possible ways,
What will you do with all those? you will mention here and
you will get the answer.
So, the first one which will be the smallest, that would be
one digit integers.
So, what are the positive integers of one digit?
So, first of all, there are one digit integers.
So, what are the possibilities of one digit integers?
One is 1, 2, 3, 4, 5.
Will there be anyone other than these?
No, there will be no one other than these.
So, how many ways are there in total?
How many are there in total?
There are 5 ways.
So, in total, we are getting 5 integers of one digit.
Next, how many are there of two digit?
How many are there of two digit integers?
So, for two digit, you have two places, one and two.
Now, in this place, you cannot put zero.
Because, as soon as you put zero in the tenths place, then
it will remain only of one digit.
But, the one digit ones have already been used.
We have removed them.
So, here you can put anything other than zero.
You cannot put zero other than zero.
You can put 1, 2, 3, 4, 5 other than zero.
You can put 5 of these.
So, how many ways are there to fill this place?
There are 5 ways to fill this place.
How many ways are there to fill this place?
Now, here you can put zero.
So, you can put zero.
Now, here one letter or one digit has been used.
So, if you don't want to repeat, then you cannot use that
letter here.
You cannot use that digit here.
So, suppose we had selected 1 here.
So, apart from 1, we can use the remaining 5 digits here.
So, what happened? 5 ways.
Now, we have to combine these two and make one integer.
These cases are together.
So, what will happen to these two?
They will multiply each other.
So, we have got 5 into 5 which is 25 integers.
This is the total number of integers.
How many integers are there?
Similarly, we will do for 3 digits.
For 3 digits, we have 3 places.
1, 2, and 3.
Zero cannot come in 100 places.
Because, it will become 2 digits.
So, apart from zero, the remaining 5 digits can come here.
So, 5.
Now, here zero can come.
And, the digit which has been used cannot come here.
So, these are also 5 possibilities.
Now, how many will come here?
See, one word has been used here.
One has been used here.
Now, how many digits are left?
From which we have to select?
In total, we had only 6 digits.
So, 2 digits have been used out of 6.
So, only 4 digits are left.
So, we can use this and fill it with 4 ways.
Now, this is one word.
It is working simultaneously.
So, what will happen to all these?
They will multiply each other.
25 into 4 is 100.
So, we will get 100 integers.
3 digit integers.
Using these 6 integers.
And, zero.
Zero also.
How many 4 digit integers will be made? 4 digit integers.
Similarly, 5 digit integers will be made.
And, 6 digit integers will be made.
So, for 4 digit, 4 digit.
1, 2, 3, 4.
So, 5.
First, zero cannot come here.
Then, zero can come here.
So, 5.
Then, it cannot be repeated here.
4.
And, 3.
So, 100 into 3 will be 300.
Similarly, 5 digit will have 5 places.
And, in this starting place, zero cannot come.
So, if zero does not come, we can fill it with 5 ways.
Similarly, we can fill it with 5 ways.
Each one will be reduced.
So, 300 into 2 will be 600.
600 total integers will be made.
Of 5 digits.
Then, for 6 digits, 1, 2, 3, 4, 5, 6.
So, 5 for first place.
Then, 5 for second place.
Then, 4, 3, 2, 1.
So, 1 is being multiplied in 600.
So, 600 will be the answer.
So, this is 1 digit, 2 digit, 3 digit, 5 digit, 6 digit.
From 1 digit to 6 digit, all positive integers using these
6 integers without repetition.
Now, we had to find the total.
So, either it will be 1 digit, or 2 digit, or 3 digit.
So, these are all your options.
So, what should I do with all these?
I will add them.
The final answer will be the number of positive integers.
When you add all these, 6 plus 6 will be 1200.
1200 plus 400 will be 1600.
1630 will be the total number of positive integers.
Read this solution once.
See if you are able to understand.
Moving on to next question.
See this.
See this question.
In a party, every person shakes hands with every other
person.
If there are 105 handshakes, find the number of persons in
the party.
After reading this question, you are not able to understand
the meaning of permutation.
You are not able to understand.
This is also a question of that field.
You are not able to understand.
But, this is a question of permutation combination.
Now, you have a party.
You have to find the number of persons in this party.
Find the number of persons in this party.
Let's say you have X persons in this party.
How many persons are there?
X persons in this party.
What is the total number of persons? X persons.
Now, you have to find the number of handshakes.
What is the total number of handshakes? 105 The total
number of handshakes is 105. Whenever there is a handshake,
how many persons are used?
Now, I am doing it alone because I am giving you an example.
But, if I do it with someone else, I am used.
And, the second person I am shaking hands with is used.
So, two persons are used in one handshake.
Similarly, the total number of handshakes is 105.
So, you are a total person.
N is the total number of objects.
X is the total number of persons.
You are choosing two persons for one handshake.
Only then, one handshake is possible.
So, you are choosing XC2.
Now, the order will not matter.
Obviously, the formula will be C.
XC2 Whose number should be equal to this? This should be
equal to this.
If I shake hands and my friend shakes hands, or my friend
shakes hands with me, both are same. The order will not
matter.
Therefore, the formula is C.
Now, the total number of handshakes is equal to 105.
The total number of handshakes is equal to 105.
What was the question?
What did we convert?
Now, we will solve this.
How will we solve?
X factorial
X minus 2 factorial into 2 factorial
Multiply 2 factorial by 105.
0, 1, 1, 2, 210
Now, take X factorial back.
2 steps will cancel out.
This is X minus 1 into X minus 2 factorial.
2 steps back.
X minus 2 factorial cancels out.
We got X square minus X is equal to 210.
This is quadratic.
Solve this.
You will get two values of X.
One positive and one negative.
Exclude negative value.
Discard negative value.
Because negative can't be persons.
X gives total number of persons.
So, it can't be negative.
Assume positive answer.
What will be the answer?
Take quadratic from this formula.
Minus b plus minus under root b square minus 4ac by 2a.
So, this becomes minus b that is minus 1 plus 1 plus minus
under root 1 square minus 4 into
This will go here and become minus, right?
So, this minus is equal to 0.
So, value of c is also minus. So, this becomes plus.
4 into 210.
4 0's is 0, 4 1's are 4, 8 40.
So, this will be 841 by 2.
So, whose root will be 841? Whose square is 841? See.
Who is 1 followed by? 11 and whose will be this? 1000,
means 10.
15, 16, 17.
Take 19 into 19.
So, this would be 81.
9 1's are 9.
17 into 19.
1.
You can solve this.
Whatever will be the value of xc.
Because it will take time to calculate.
You can comment.
Then we will verify the answer.
Now, this class is getting long.
So, let's do the last question quickly.
See the last question.
From 6 men and 4 ladies, a committee of 5 is to be formed.
How many? 6 men and 6 ladies.
A committee of 5.
How many committees do we need?
A committee of 5.
Means, we have to choose 5 from these.
In how many ways can this be done?
If the committee is to include at least 1 lady.
At least 1 female should be there.
So, there are 5 places.
We have to choose 5 people.
Now, how can you choose these 5 places?
See, there are 5 places.
So, at least 1 lady should be there.
Means, 1 lady should be used here.
1 female should be used here.
Then, you can put anything here.
Either put men.
Or put female.
So, you will get the cases.
The first case is this.
The first case has 1 F.
And the rest are males.
The second case is this.
There are 2 females.
This case has 1 female.
And the rest are 4 males.
The second case has 2 females.
And the rest are 3 males.
The third case has 3 females.
And 2 males.
The fourth case has 4 females.
And 1 male.
And the fifth case has 5 females.
So, these are the corresponding cases.
Now, you have to multiply these individual cases.
And you have to add these cases to these individual cases.
So, let's see how to do it.
First, see.
1 female and 4 male.
So, you will select 1 female out of 6 females.
So, you will get the first case.
You are selecting 1 from 6.
You will select 2 from 6.
Because this is the same case.
1 female and 4 male.
So, you will select 2 from 6.
You will select 4 from 6.
So, you will get the first case.
The second option is 2 female and 3 male.
So, you will select 2 from 6.
You will select 2 from 6.
You will select 3 from 6.
The third option is 3 female.
So, you will select 3 from 6.
You will select 2 from 6.
The fourth option is 4 female from 6.
You will select 1 male from 6.
You will select 1 from 6.
The fifth option is 5 female from 6.
You will select 5 from 6.
Now, you have to add these cases to these individual cases.
So, you will get the final answer.
We will add all these cases.
So, I hope you have understood the questions of Permutation
and Combination.
You can solve these questions in exam too.
So, we will meet in the next video.
Till then, you can practice these questions.
And, thank you for watching the video.
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