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0:12
Hello students! I am Dr Gajendra Purohit and you're watching my youtube channel where I upload videos on engineering, BSc and mathematics student if you're preparing for a competitive exam where general aptitude is asked then my channel could be really helpful i've uploaded many videos on general aptitude, to watch them click on the i tab today i'll be discussing an important topic that is permutation
0:58
Permutation simply means arrangement of the numbers or things
1:11
lets take an example
1:17
lets say we have 3 persons A, B and C and we need to arrange them
1:26
we'll fix A at the 1st position and arrange B and C
1:40
we'll fix A at the 2nd position and arrange B and C we'll fix A at the 3rd position and arrange B and C so that way we have 6 arrangements this was a 3 letter arrangement so we did it quite quickly but for more values we use permutation
2:33
I'll give you a trick to use to in the exams
2:47
we have 3 letters and we need to arrange all the 3 letters
3:04
the trick we have is we need to write 3 backwards thrice like 3*2*1=6
3:22
for this we write 3 backwards twice
4:08
now lets arrange 2 out of the 3
5:31
A student usually gets confused between combination and permutation
6:20
there are three letters A,B and C and you need to choose any two combination is selection while permutation is selection + arrangement
6:36
the only difference here is that in combination we are not concerned with the sequence
6:56
so here the answer is 3 because in combination sequence doesn't matter
7:06
but in permutation we consider the sequence as well
7:18
combination is selection while permutation is selection + arrangement
8:30
the concept of repeated and non repeated is very important
8:48
a digit number needs to be formed with no digits being repeated
9:00
so for the 1st position to fill we have 5 options now after putting one of the numbers at the 1st position we are left with 4 numbers
9:25
now we are left with 3 numbers similarly we have 2 and 1 number left
9:42
now using the formula we need to choose 5 letters out of 5 letters we need to write 5 backwards 5 times
10:36
now if the digits repeated then digits repeating means all the numbers can take all the places in the previous concept we could not repeat the numbers at different positions
10:58
whenever the digits can repeat every position has 5 options as the digits can repeat
11:24
here we won't use the concept of permutation
11:54
in this question we are not told that whether it is repeating or non repeating so for such question we use non repeating always
12:22
now for 5 digit the first position cannot be taken by 0 as then it will be a 4 digit number
12:35
for the first position we can only keep 4 options out of the 5 as if 0 takes the first position then it'll be a 4 digit number
12:50
but now 0 can take up the second position
13:06
so for the first place we have only 4 options 2,4,6 and 8 now for the first position one number is fixed and we have 4 numbers left as 0 can now take the rest of the positions
14:04
we need to form 5 digits
14:23
but the first position cannot be taken up by 0
14:31
so we have only 6 options to fill the first place
14:40
we have fixed one number at 1st position and now rest of the numbers can take the 2nd place
14:51
now we reduce the numbers as the numbers cannot be repeated
15:35
this is. a tricky question, and you need to learn the trick here
15:48
for it to be a even number it should end in 0, 2, 4,6
16:04
now if we fix 0 at the end
16:12
now we have 6 options left so the 1st place has 6 options and the 2nd place 5 options the ways to fix the 0 was 1
16:32
it does not mean that you multiply with 0 , the number of ways a zero can be fixed is 1 so 6*5*1
16:46
now in the last place only 2,4 and 6 can be placed as we need a even number so we have 3 ways to fill the last digit
17:07
another condition is that the first position should not be taken up by 0 so now the number of ways we have is 5 as we have fixed one number at the end and 0 can't take 1st position now one number has taken 1st place and another has taken the last place but now 2nd place can be taken by 0 as well so we have 5 ways out of 7 so we add the total ways of getting an even number 30+75
18:05
whenever we are asked to form even numbers then we solve for 0 and other even numbers separately
19:18
which means no repetition
19:35
greater than 800 so from 801 to 999
19:47
next will be from 1000-3999
20:02
first we check the 3 digits numbers
20:10
when we talk about 801-999 we can only take 2 numbers at the 1st place i.e. 8 &9 so we two ways to fill the 1st position now the 2nd place can be filled with 7 numbers so 7 ways
21:00
now foe this 4 digit number 1st place can be filled with 1,2 so 2 ways and the rest of the places can be filled with any of the numbers now as we have taken 1 of the numbers already we have 7 left then 6 and then 5
21:48
we add the total ways
22:00
i hope you must have learned the concept very well in the next class i'll be teaching the concept of circular permutation the concepts and tricks to arrange people around a round table and the concept of necklace
22:25
stay tuned for more such content and support the channel Like, Share and keep commenting
22:44
this is my channel that I'm working on this is my channel for higher mathematics for engg and BSc
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