1:47:33 The question asks that how many words are listed in dictionary when we open the dictionary before E letter. Means we have to tell all the words coming before letter E no matter what is the meaning without repetition other than given in the word " EXAMINATION" . @ushakghai sir
1:47:33 H.W. Q. = We have to put the first letter A and find the permutations of other 11 letters , so that we can find all the words that list before the first letter E. ❤
1:47:33Answer The letters of the given word are A, A, E, I, I, M, N, N, O, T, X,i.e, Word starting with A are formed with the letters 2 I's, 2 N's, A, E, X, M, T, O (total 10 letters). Hence, number of words formed by these letters =10! / 2!2! =10×9×8×7×6×5×4×3×2×1/ 2×1×2×1 = 10×9×8×7×6×5×3×2×1/ 4 = 907200 ways
I think some of you must have thought about it once in your life that I used to think, How can there be so many phone number with just 10 numbers? And now I know that there can be 10^10 ways to write a phone no. which is equals to 10 billion where Earth's population is 8 billion, which means even if we give each person on Earth a different phone number we still have 2 billion different phone number we can give 🤯😵😂
1:47:33 If the different permutations of all the letter of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with E? Answers: Alphabetically only A before E in the letters of word EXAMINATION So only words starting with A letter befor the the first letter starting with E letter .•. number of ways = (AA ) EXMINTION thereis 10 leters but A is always 1st .•.9! In this 9 letters I & N is repeating 2 times so 9!/2!•2!
@@yogalakshmi-fq9vxThe unit place can be filled only by 2 numbers I.e., 2,4 Therefore permutations of unit place = 2 P 1 Also, now the 3 places can be filled by 4 no I.e., 1,3,5,(either 4 or 2) because 1 of them is already used Therefore permutations of other 3 places will be= 4 P 3 Also net permutations= 4 2 P × P 3 1 =4! ×2! =48
1:47:30 ans of words with using word ‘EXAMINATION’ that are in dictionary before the first word starting with E are 9,07,200 A_ _ _ _ _ _ _ _ __ Using 10!/2!2!
54:00 total numbers will be 120 out of which 48 are even ❤😊 1:17 cases when I not come together is 33810 1:47:00 sir this question means that we have to arrange all the words in dictionary pattern first letter will start from A and then so on 1:15:04 ans1 is 420 ways ans 2 is 778320
The letters are arranged in a dictionary in alphabetical order. Therefore, in this question, we have to find the number of words starting with A, B, C, D to get the number of words before the first word starting with E appears. Therefore, We first try to find the total number of words starting with A. For this, we fix A at the beginning and find the different permutation of the other letters after that. As the total number of letters in EXAMINATION is 11, the words starting with A should be of the form A _ _ _ _ _ _ _ _ _ _ Now, after taking out A, the remaining letters are E, X, M, I, N, A, T, I, O, N. We know that the number of distinct permutations of n objects (having k distinct objects repeated one or more times) in which object 1 is repeated p1 times, object 2 is repeated p2 times etc. is given by n!p1!p2!p3!...pk! ....................(1.1) However, we note that of these remaining letters, there are two I’s and two N’s which are identical. Thus, using equation (1.1) the number of distinct permutation of the remaining letters are 10!2!2!=907200 ..............(1.2) Thus, the number of words from the letters of EXAMINATION are 907200. As B, C and D are not present in the letters of the word EXAMINATION, the total number of words in the list before the first word starting with E is equal to the number of words starting with A, which is 907200. Note: In this question it is necessary to divide 10! by 2! and 2! in equation (1.2) because the words in which the two I’s or two N’s would be interchanged would correspond to the same word in the dictionary.
1:47:33 sir iska matlab hai ki , humara jo word examination hai ,uske sare permutations are diff , toh Hume number of arrangements batani hai , for jitne bhi words "E" se phele ayenge , because dictionary main words A-Z arrange hote hai
53:54 SIR THIS QUESTION HAVE TWO ANSWERS 1-[60 ways] 2-[90 ways] I CAN EXPLAIN IT. agar mene 1st digit 2 leya to last box me sirf 2no. hi rejayege AND agar mene 1st digit 3 choose kiya phir mere pe last box me 3 no. bachenge
@@ShreyaKumari-b9j But I think it will be 39916800/1152=34650 And occur together will be 1680/2= 840 Therefore, 34650-840= 33810. Main ans...thank me later 😅
1:47:13 the question is asking the possible arrangement of word EXAMINATION starting from letter A ( Because it is the only letter that comes before E starts in dictionary from the given letters in examination ) which is given by= 10! /(2!*2!) .
1:47:18 isme A ko aage rakhke baki words ko box me dalke fir arrange krna h mtlb factorial form me or jo reapet h use upon me kr dena h too A se bnne wale words pta chl jayenge or E se pehle A ke alawa koi dusra word nhi aata
sir sabse phle iske andar hm examination word ke permutation nikaalenge then uss word ke nikaalenge jb e first letter hoga usko freeze karke then total waale mein se minus karenge e waale ko answer 907200 arha h same tareeka jaise apne ek question phle bhi karaya tha star and triangle wala.......
In this question we arrange the letter by alphabetically the question say that we arrange the word examination which is listed in dictionary and the word starts from E we arrange all letter except E by alphabetical order and find the no. Of words Thank you❤
21:43 ---- Ans 50,40 24:30 ---- Ans 366 37:51 ---- Ans 320 1:47:33 H.W. Q. = We have to put the first letter A and find the permutations of other 11 letters , so that we can find all the words that list before the first letter E.
Sir work power and energy and system of particles ki video lado sir please sir half yearly anae vale hai mai aap ki video ka intezar kar raha hu mujhe asshu sir se hi physics samaj aati hai
it was a lecture of 2 hours and i completed this lecture in 5 hours it have so many important questions i made notes and wrote allthr questions in the notebook, but lecture was worth 5 hours
1:47:45 To find the number of permutations of the word "EXAMINATION" that start with a letter other than "E", we can consider the remaining letters after fixing "E" at the beginning. The word "EXAMINATION" has 11 letters in total, including 3 "I"s, 2 "A"s, and 2 "N"s. If we fix "E" at the beginning, we have 10 positions remaining for the other letters. Out of these, there are 3 positions for "I", 2 positions for "A", and 2 positions for "N". The remaining 3 positions can be filled with the letters "X", "M", and "T". So, the total number of permutations before the first word starting with "E" is \( \frac{10!}{3! \times 2! \times 2!} \). Calculating this: \[ \frac{10!}{3! \times 2! \times 2!} = \frac{3628800}{12 \times 2 \times 2} = \frac{3628800}{48} = 75600 \] Therefore, there are 75,600 words in the list before the first word starting with "E".
1:47:33 isme pehle alphabetical letter fix krenge ex A ko fix kia phir bhakiyo ko n! tarike se arrange krenge orr tab tak krenge jab tak hum E ko fix na kr de
the lecture by Ushank sir is a complete brief of the rd+ncert book if we watch these lectures and solve RD question it will feel like a piece of cake and ncert also covered with it and ya I tried Jee Mains level questions also the after lecture i was able to solve 80-90% questions from rd sharma and modules correctly sir we want a complete series for 12th and 11th asap thank you sir for always helping us I hope we get jee mains and advanced level content of full course too
1:48:12 We want the words which come before words which started with E so we take those alphabets which comes before E That is only A So we fix the letter A at first position and the arrange remaining letters by permutation
Using the letters of the word "EXAMINATION", without using any letter more that the times it is actually used in the actual word, we have to find out how many words can be there in the dictionary before words of 'E' as starting. Since only 'A' is a letter before 'E' that is used in the word, we have to find out the number of all words starting with 'A' upon permutating. Answer: total number of letters = 11 fixed letter = 'A' no. of letters available for permutating = 10 no. of letters repeating (a is not considered repeated here since one 'a' is already fixed) = 2, where both are repeated twice. So, total number of words = 10! / (2! * 2!) = Calculate yourself!!
Hello bacho event like karo jaldi jaldi !
❤❤❤
hello
Hello bachchooooooo
Ye chapter tereko smjhna nhi aaya bro
😂❤❤❤
0:00 Introduction
2:50 Fundamental principles of Counting
27:02 Factorials
39:09 Permutation
1:22:47 Combination
Thanks
Thanku
Thanks
Thanks a lot
😊
1:47:33 The question asks that how many words are listed in dictionary when we open the dictionary before E letter. Means we have to tell all the words coming before letter E no matter what is the meaning without repetition other than given in the word " EXAMINATION" . @ushakghai sir
10:31 3x2x5 = 30 different ways
11:09 115x45x16 = 80800 ways
22:47 10 digits i.e. 0,1,2,3,4,5,6,7,8,9
115x45x16=82800
@@vkx_editx bhai badiya kiya aapne
54:15 Total - 120, Even - 48
How😢
1:47:33 H.W. Q. = We have to put the first letter A and find the permutations of other 11 letters , so that we can find all the words that list before the first letter E. ❤
I think we have to take 10 other letters to arrange
0:00 Indroduction
2:50 Fundamental principles of Counting
27:02 Factorials
39:09 Permutation
1:22:47 Combination
1:47:33Answer
The letters of the given word are
A, A, E, I, I, M, N, N, O, T, X,i.e, Word starting with A are formed with the letters 2 I's, 2 N's, A, E, X, M, T, O (total 10 letters).
Hence, number of words formed by these letters
=10! /
2!2!
=10×9×8×7×6×5×4×3×2×1/
2×1×2×1
= 10×9×8×7×6×5×3×2×1/
4
= 907200 ways
Hey your answerscwrong in den9minator you have to take one more 2 factorial
Yesss@@JagatSinghYadav-x9n
@@JagatSinghYadav-x9n why?
54:10
hw: 120 normal and 48 in even
0:00 Indroduction
2:50 Fundamental principles of Counting
27:02 Factorials
39:09 Permutation
1:22:47 Combination
I think some of you must have thought about it once in your life that I used to think, How can there be so many phone number with just 10 numbers? And now I know that there can be 10^10 ways to write a phone no. which is equals to 10 billion where Earth's population is 8 billion, which means even if we give each person on Earth a different phone number we still have 2 billion different phone number we can give 🤯😵😂
moreover there are country codes also..
Abe gawar 💀🤡🤡
1:47:33 If the different permutations of all the letter of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with E?
Answers:
Alphabetically only A before E in the letters of word EXAMINATION So only words starting with A letter befor the the first letter starting with E letter
.•. number of ways =
(AA ) EXMINTION
thereis 10 leters but A is always 1st
.•.9! In this 9 letters I & N is repeating 2 times so 9!/2!•2!
Sorry to correct you but your answer is wrong in the place of 9! It would be 10!
Its also verified in ncert
.
Timestamp:- 1:12:38
All vowels occur together= 240
Exactly
54:10 Total ways 120 and even numbers 48
How 48🙃
@@yogalakshmi-fq9vxThe unit place can be filled only by 2 numbers I.e., 2,4
Therefore permutations of unit place = 2
P
1
Also, now the 3 places can be filled by 4 no I.e.,
1,3,5,(either 4 or 2) because 1 of them is already used
Therefore permutations of other 3 places will be=
4
P
3
Also net permutations=
4 2
P × P
3 1
=4! ×2!
=48
Bruh there is no six in the question bro😅@@princebhardwaj-kb8vf
@@Defeatm1943Mno
@@Defeatm1943M great work!
1:51:05 Me , Akul and Lavi are those 3 students❤️🔥💕
1:16:46
Total no of ways- 34650
Ans - 33180
33810 hoga
1:12:36 MONDAY - all vowels occurred together - 240 ways
exactly!!
1:47:30 ans of words with using word ‘EXAMINATION’ that are in dictionary before the first word starting with E are 9,07,200
A_ _ _ _ _ _ _ _ __
Using 10!/2!2!
No it will be 10!/2!.2!.2!
1:25:45
Sir moka to do icecream choose krne ka😂😂tab na karenge😅......no hate to sir i really like the way he teaches ❤❤❤😊😊😊
54:00 total numbers will be 120 out of which 48 are even ❤😊 1:17 cases when I not come together is 33810 1:47:00 sir this question means that we have to arrange all the words in dictionary pattern first letter will start from A and then so on 1:15:04 ans1 is 420 ways ans 2 is 778320
But how 48 are even
@@ShauryaJain-kz1lk total outcomes are 120 out of which 48 outcomes are even
@@ShauryaJain-kz1lk4×3×2×2
Your answer at 1:15:04 is wrong
Ans 1 is 34650
Ans 2 is 840
Final answer is 33810
@@aaditjain5051pls check 🙏 what I have written 📝 🙂
54:12 Normal Answer =120 , even answer=48
How?
5*4*3*2*1=120
4*3*2*2=48
The letters are arranged in a dictionary in alphabetical order. Therefore, in this question, we have to find the number of words starting with A, B, C, D to get the number of words before the first word starting with E appears.
Therefore,
We first try to find the total number of words starting with A. For this, we fix A at the beginning and find the different permutation of the other letters after that.
As the total number of letters in EXAMINATION is 11, the words starting with A should be of the form A _ _ _ _ _ _ _ _ _ _
Now, after taking out A, the remaining letters are E, X, M, I, N, A, T, I, O, N.
We know that the number of distinct permutations of n objects (having k distinct objects repeated one or more times) in which object 1 is repeated p1
times, object 2 is repeated p2
times etc. is given by n!p1!p2!p3!...pk! ....................(1.1)
However, we note that of these remaining letters, there are two I’s and two N’s which are identical. Thus, using equation (1.1) the number of distinct permutation of the remaining letters are 10!2!2!=907200 ..............(1.2)
Thus, the number of words from the letters of EXAMINATION are 907200.
As B, C and D are not present in the letters of the word EXAMINATION, the total number of words in the list before the first word starting with E is equal to the number of words starting with A, which is 907200.
Note: In this question it is necessary to divide 10! by 2! and 2! in equation (1.2) because the words in which the two I’s or two N’s would be interchanged would correspond to the same word in the dictionary.
😊
vedantu se copied lol
😂@@MahirajSinghRathore-vf8vg
Any Pluto wasi here 👽
yes
Yes
ARJUNA JEE 'Tarun' sir ka dialogue
yes
Yes
54:14 i)5!=120 and ii) 2P1 X 4P3 = 2! X 4! = 48
39:25 Permutations means arrangement of something like A word,letter,number,etc
Hnn you are right
1:47:33 sir iska matlab hai ki , humara jo word examination hai ,uske sare permutations are diff , toh Hume number of arrangements batani hai , for jitne bhi words "E" se phele ayenge , because dictionary main words A-Z arrange hote hai
YES JANN
Like if u have exam tomorrow 😂😂😂
How was the exam? Did this oneshot worked?
( I have my exam tomorrow)
Yes brother I want it @@AMANSEP
Bro i have 😂
Compartment ka hai
@@btrgaming5965😂 same
54:11 ans 5p4 = 5!/(5-4) = 5×4×3×2×1=120
4p3= 4!/(4-3)= 4×3×2×1=24
53:54 SIR THIS QUESTION HAVE TWO ANSWERS 1-[60 ways] 2-[90 ways] I CAN EXPLAIN IT. agar mene 1st digit 2 leya to last box me sirf 2no. hi rejayege AND agar mene 1st digit 3 choose kiya phir mere pe last box me 3 no. bachenge
Bhai same pov par mujhe pata h ki kisi teacher ko bolunga to vo Bina kisi explanation ke mujhe galat bol denge
54:09 No. Of 4-Digit no. = 120
No. Of 4-Digit Even No. = 48
Permutation=arrangement is important
Combination=arrangement is not important
Thanks bro
@@vijjiworld My pleasure 😊
@@reshmapathan8593 bro now you are in which class
Thank you sir this chapter is very important for ours
ending aate hue saare difficult aur important wale questions toh aapne home work de diye sir
Plz sir difficult question kariye 😢😢 49:25
Thank u sir ,aaj hi pura lecture dekh liya or samajh liya hai ......aiase hi padhate jaoo.... abhi extra question laga raha hu ...
Aree hardik Bhai tussi idhar ke karr rahe ho 😂
@@utkarshshukla2219 bus Bhai aapki cripa hai 😂😂
Me bhi aa gya 😎😎
Swagat nahi karoge humara!?
@@Pranjwalkumar aare kyo nahi Bhai
1:47:25 question says that what are the total number of words that can be formed with this word starting from E
10!/2!2!2!
Today is my birthday and math exam😢.... 19 oct 2024...5.02 am....❤
sorry bro late ho gya video dakna ma happy birthday
54:07 no.of ways=120
FOR EVEN NO . = 48
1:15:03 1663200/4×3×2×2 = 4989600
Thnx I'm finding the for few minutes finally I found it
@@ShreyaKumari-b9j But I think it will be 39916800/1152=34650
And occur together will be 1680/2= 840
Therefore, 34650-840= 33810. Main ans...thank me later 😅
1:47:13 the question is asking the possible arrangement of word EXAMINATION starting from letter A ( Because it is the only letter that comes before E starts in dictionary from the given letters in examination ) which is given by= 10! /(2!*2!) .
1:40:49 Daughter wale question mai permutation use hona chahiya tha quki block bhi apna under rearrange hosakta hai
Mid term aur periodic test m rat kr kra tha ye chapter, aaj phli baar samajh aaya h ,even miscellaneous ke ques bhi khud ho rhe h
Bhai samjh nahi ara
Sahi hai bhai
1:48:12 sir humein order me create karna hain as in a dictionary so we need all words starting with A upon number of repetition.
1:47:18 isme A ko aage rakhke baki words ko box me dalke fir arrange krna h mtlb factorial form me or jo reapet h use upon me kr dena h too A se bnne wale words pta chl jayenge or E se pehle A ke alawa koi dusra word nhi aata
Red
can be written as
In 6 ways as red
Der
Erd
Dre
Rde
Edr
Best lect of PnC on whole TH-cam fr❤👍 thanks a lottt Sir✨️
1:48:23 Rank of word in the dictionary 😊😊
Thank you sir for the great lecture ...😊
sir sabse phle iske andar hm examination word ke permutation nikaalenge then uss word ke nikaalenge jb e first letter hoga usko freeze karke then total waale mein se minus karenge e waale ko answer 907200 arha h same tareeka jaise apne ek question phle bhi karaya tha star and triangle wala.......
54:14 Normal - 120
Even - 48
👍
Legends are watching this just 2 hours before half yearly exams
Us 😅😂
😂
Me
Us @@Miyuki-OtakuSensei
Bhai yeh one shot se kuch fayda hua?
11:17 Girls can get ready in 82,800 wayss
In this question we arrange the letter by alphabetically the question say that we arrange the word examination which is listed in dictionary and the word starts from E we arrange all letter except E by alphabetical order and find the no. Of words Thank you❤
not except E but letter before E in alphabatic order (ABCD)
❤
1:41:48 sir yeh 5 factorial ka concept samajh nhi aya.
Sir mid term ki tayaari shuru karva do
54:15. Normal answer-120
Even answer-48
54:13 , 120 total no. Of ways and 48 even no. Of ways
21:43 ---- Ans 50,40
24:30 ---- Ans 366
37:51 ---- Ans 320
1:47:33 H.W. Q. = We have to put the first letter A and find the permutations of other 11 letters , so that we can find all the words that list before the first letter E.
54:15 Total mein 120 but even wale mein 48
sahi hai kya??
Sir work power and energy and system of particles ki video lado sir please sir half yearly anae vale hai mai aap ki video ka intezar kar raha hu mujhe asshu sir se hi physics samaj aati hai
1:48:19 We have to multiply 10!/2! Four times
54:00
120 total nos.
48 even
1:42:50
2!x3!x5!
1:56:58
SIR WE WANT LIVE CLASS 🙏🙏 FOR CLASS 11 MATHS CHAPTERS.
🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥
it was a lecture of 2 hours and i completed this lecture in 5 hours it have so many important questions i made notes and wrote allthr questions in the notebook, but lecture was worth 5 hours
I completed the same in 45 minutes...
1:47:45
To find the number of permutations of the word "EXAMINATION" that start with a letter other than "E", we can consider the remaining letters after fixing "E" at the beginning.
The word "EXAMINATION" has 11 letters in total, including 3 "I"s, 2 "A"s, and 2 "N"s.
If we fix "E" at the beginning, we have 10 positions remaining for the other letters. Out of these, there are 3 positions for "I", 2 positions for "A", and 2 positions for "N". The remaining 3 positions can be filled with the letters "X", "M", and "T".
So, the total number of permutations before the first word starting with "E" is \( \frac{10!}{3! \times 2! \times 2!} \).
Calculating this:
\[
\frac{10!}{3! \times 2! \times 2!} = \frac{3628800}{12 \times 2 \times 2} = \frac{3628800}{48} = 75600
\]
Therefore, there are 75,600 words in the list before the first word starting with "E".
not 3 "I" its only 2
@@admiralscott4119 ok thanks
Galat hai lod* answer 907200 hai 🍆
Sir actually sequence and series is also coming in mid terms so please bring a lecture on it too......😢
No bro
THIS LESSON COVERED ALL THE TOPICS OF NCERT ,,, THANK U SIR FOR THIS LECTURE , HOPE YOU WILL BRING SOME IMPORTANT QUESTIONS OF THIS CHAPTER
1:37:53 Dammnnnn bro.Not gonna lie man my cycle used to have same lock no. 💀💀💀😳😳
😂😂
C stands for choose
So whenever you had to choose or select use combination
My name is Adi Pandey so..A stands for arrangement and P for permutations
🧠🧠
A-an
p-padh
anpadh
25:53 sir isme 5×3 hona chaiye na ? Kyunki 2 flags toh use hogya .
Sir plz mid term ke liyeee marathons upload kr doo😭😭😭😭😭😭😭😭😭😭😭😭
Tere mkc
Best lecture for one day before exam😊
1:31:23 Sir common jo nikala hai vo glt nhi hai ?
Same dougt
Sahi h bro
1:47:33 isme pehle alphabetical letter fix krenge ex A ko fix kia phir bhakiyo ko n! tarike se arrange krenge orr tab tak krenge jab tak hum E ko fix na kr de
Sir voice is similar to Ashu sir
Both are brothers. Exam kab hai? Mera kal hai 😢
5:50....sir giving most important lesson of this chapter
Kis kis ka exam 19 oct ko hai 😂
Mera aaj hai
Mera
Bhai tera kounsa school hai
us bhai
Delhi Government 🙂
1:16:41 33810=permutation of mississippi I do not coming togather
Fantastic lecture 👍
Thanks❤
1:51:20 adamyaaa harshhh krishnaaa
5:55 bro got emotional 😂
Thank u sir jarurat thi iski 😢.....aur bhi chapter thoda jaldi laiye sir pls 🙏
Acha 🙁 sachi batti Or hamro jarurt na h kya 😂
@@mukulsharma4409 aukat mein reh... sir tera nokar nhi hein ... tu bachi hein aukat mein rahiyo
54:16 Normal -----> 120 [5!]
Even -----> 48 [4! × 2!]
Where is my heart ❤️ 😑 ?
in your body
bro this video is 11 months old
@@anilguptta1908 yt heart brother 💀😭
@@-ar_yan well I know that 🥹
1:12:40 - all vowel occur together= 240
Bhut jyadaa jarrurat thi sirrrr🥺🥺🥺🥹 thanks a lotttttt❤❤❤❤
the lecture by Ushank sir is a complete brief of the rd+ncert book if we watch these lectures and solve RD question it will feel like a piece of cake and ncert also covered with it and ya I tried Jee Mains level questions also the after lecture i was able to solve 80-90% questions from rd sharma and modules correctly sir we want a complete series for 12th and 11th asap thank you sir for always helping us I hope we get jee mains and advanced level content of full course too
Fake
Faiku
Abe pheku ye bas ncert level ke questions hai
Matlab kuch bhi
Kitne paise liye?
54:09 even no 48 and total no formed 120
Sir iske baad sequence and series ka Lana please
Gendu
1:36:38 ncr imp concept
Live class vote for class 11👉👉👉👉👉
Live class
1:48:12 We want the words which come before words which started with E so we take those alphabets which comes before E
That is only A
So we fix the letter A at first position and the arrange remaining letters by permutation
It's 7am right now. My exam is at 9am . My bus is at 8am . Am brushing my teeth
Pls pee on your face for this extra comment 🎉
54:10
i) 120 ways
ii) 48 ways
Using the letters of the word "EXAMINATION", without using any letter more that the times it is actually used in the actual word, we have to find out how many words can be there in the dictionary before words of 'E' as starting. Since only 'A' is a letter before 'E' that is used in the word, we have to find out the number of all words starting with 'A' upon permutating.
Answer: total number of letters = 11
fixed letter = 'A'
no. of letters available for permutating = 10
no. of letters repeating (a is not considered repeated here since one 'a' is already fixed) = 2, where both are repeated twice.
So, total number of words = 10! / (2! * 2!) = Calculate yourself!!
But I is repeating so the letters starting with A should be 10!/2!
I is also repeating
@@dishamathur8088 Exactly 👍
907200 is the answer 😊
@@dishamathur8088 Bhai n is also repeating
54:12
5 P 4
²P1 × ⁴P3
Sir never hesitate 😂😂
11:24
Agree 💯👍 Sir not all five fingers are same...
Like if your exam is today and your watching 3hours before😂
54:13
120 and 48
120 due to 5p4 and 48 due to 2*3*4*2=48 because digit is not repeated
35:15 sir roasted whole anti-national community just in few words 🗿🗿
Padhai me man laga😡
@@TSHUKISHIROWahi toh
Faaltu ki debate me Lage hai sab.
India=Bharat.
Padhai chhodkar baaki sab karna hai
1:48:11 we have to find the rank of the ord EXAMINATION in dictionary