Permutation and Combination One Shot | Class 11 Maths NCERT Chapter 7 by Ushank Sir Science and Fun

[@scienceandfuneducation]

Hello bacho event like karo jaldi jaldi !

[@newgammer3253]

❤❤❤

[@CODERBOYVANSH]

hello

[@LK_Rajput72]

Hello bachchooooooo

[@md.shakilwarsi6224]

Ye chapter tereko smjhna nhi aaya bro

[@navneet4861]

😂❤❤❤

[@kshitizgambhir8652]

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

[@anjaliraj2510]

0:00 Introduction 2:50 Fundamental principles of Counting 27:02 Factorials 39:09 Permutation 1:22:47 Combination

[@jatinkumar8299]

Thanks

[@SagarGurjar-i7v]

Thanku

[@ZERO-qk2zx]

Thanks

[@DhasuIdea]

Thanks a lot

[@AinooKhan-c3l]

😊

[@Ruzen911]

00:02 There are 24 ways to get ready for different combinations of clothing and accessories 02:08 Understanding permutation and combination 09:02 Understanding permutation and combination through examples 11:41 Learning permutations and combinations 19:33 Permutation can be used to arrange objects in different ways. 21:54 Understanding permutations and combinations are essential for creating phone numbers. 26:51 Factorial is a unique tool with properties. 29:23 Permutation and combination concepts explained with examples 35:53 Permutation and Combination can be solved using factorials. 38:57 Permutation involves arranging things in different ways 44:28 Permutation and Combination 47:47 Understanding permutation and combination with examples 53:26 Permutation and Combination concepts, and practical problem-solving 57:25 Understanding permutation and combination 1:03:39 Permutation and Combination of letters starting with vowels 1:06:19 Understanding permutation and combination of letters 1:12:26 Understanding the concept of permutation and combination with vowels and factorial 1:16:15 Understanding permutation and combination 1:23:37 Understanding the concept of combination and its daily applications 1:26:22 Permutation and combination are discussed with examples. 1:35:33 Understanding permutations and combinations in simple scenarios. 1:39:18 Understanding permutation and combination 1:45:34 Understanding permutations and combinations 1:49:04 Permutation and combination concepts explained 1:55:52 Arranging 5 factorials with more seats for boys

[@RakeshKumarYadav-vg4ux]

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 🤯😵😂

[@stratasphorte]

moreover there are country codes also..

[@tan1shx17]

Abe gawar 💀🤡🤡

[@anshuman7et]

Damm😊

[@hhgamers6993]

Technically speaking tho, you have to eliminate the number 0000000000 and you also have the case of 000000000 followed by any of the numbers between 1-9 00000000 followed by any of the numbers between 1-9 0000000 followed by any of the numbers between 1-9 : : : So, the actual number of possible 10 digit phone numbers are (10^10 - (1+9+9^2+9^3+9^4..9^9)) Applying GP formula within the brackets, we get (10^10 - (1(9^10 - 1)/(9-1) This simplifies to 9,564,151,950 or 9.5 billion combinations. So, yeah you'd still have enough phone numbers for every person on the planet!!

[@raghavbairagi-mz8gt]

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. ❤

[@NOT_GAMING_GUY]

I think we have to take 10 other letters to arrange

[@soulgaming6541]

Just find no of letters that can be arranged from A you'll get the answer bcoz E se phle bas A ke words aayenge total words find krne ki bhi need nahi hai

[@mechlegend2813]

54:15 Total - 120, Even - 48

[@GunjanSingh-c9h]

How😢

[@everythingthatiwanttxt5712]

​@@GunjanSingh-c9hTotal= 5×4×3×2 (kyunki repeat nhi honge)=120, even=4×3×2×2(total 5 digits hain, even bola h, toh last place aka one's place pe 2,4 hi ho sake hain for this no. to be even, mtl ki 2 digits; phir baaki bache 4,3,2 kyunki repeat nhi hone hain toh, 4×3×2×2=48.

[@AmarSingh-p2j3s]

​@@everythingthatiwanttxt5712bhai ye to thik h pr ise bina multiplication principal se kese kru ese to easy h

[@NexZ777]

​@@AmarSingh-p2j3sfor even 3p1 x 4p3 ....for normal 5p4 ...factor wula method se

[@Faahim_official_channel_1]

correct answer!!!

[@v.k_edition]

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

[@MrRegit_BGMI]

115x45x16=82800

[@PriyanshuKumar-zz4fj]

@@v.k_edition bhai badiya kiya aapne

[@ruz_20056]

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

[@JagatSinghYadav-x9n]

Hey your answerscwrong in den9minator you have to take one more 2 factorial

[@SDGAMERYT1234]

Yesss​@@JagatSinghYadav-x9n

[@himanishsaidkys]

@@JagatSinghYadav-x9n why?

[@anshgaming4097]

​​@@himanishsaidkys 2A , 2I , 2N so 2!×2!×2!

[@siddboi]

@@anshgaming4097 no bro he is right.... there will only be 2 factorial two times because on of the A is fixed at the start

[@Sam.69.]

0:00 Indroduction 2:50 Fundamental principles of Counting 27:02 Factorials 39:09 Permutation 1:22:47 Combination

[@yush2007]

54:10 Total ways 120 and even numbers 48

[@yogalakshmi-fq9vx]

How 48🙃

[@Defeatm1943M]

​​@@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

[@Defeatm1943M]

Bruh there is no six in the question bro😅​@@princebhardwaj-kb8vf

[@animewatcher400]

​@@Defeatm1943Mno

[@Maxverstappen1812]

@@Defeatm1943M great work!

[@mihiracharya1361]

1:47:25 question says that what are the total number of words that can be formed with this word starting from E

[@mihiracharya1361]

10!/2!2!2!

[@rajeevpandey6220]

Any Pluto wasi here 👽

[@shadow19113]

yes

[@MayankPathak-jc5oi]

Yes

[@ramkumarsingh3389]

ARJUNA JEE 'Tarun' sir ka dialogue

[@vishwakarma368]

yes

[@YashwinVashishath-x3s]

Yes

[@gunulivevlogs12]

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!

[@Zoro.king4087]

No it will be 10!/2!.2!.2!

[@mugilnilalakshmi3594]

​@@Zoro.king4087 bro no.. its 907200 only.. because see two A's are not included in the denominator as it is already been calculated as single alphabet together with other alphabets that is 10!.. now for denominator we write 2! ( N repition 2 times ) another 2! ( I repetition 2 times) but not A repition two times as A's repition is already included in the 10!

[@kajalsen77]

Timestamp:- 1:12:38 All vowels occur together= 240

[@golluyadav8247]

Exactly

[@manasranjan4330]

54:10 hw: 120 normal and 48 in even

[@1deepakx]

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

[@ShauryaJain-kz1lk]

But how 48 are even

[@1deepakx]

@@ShauryaJain-kz1lk total outcomes are 120 out of which 48 outcomes are even

[@OvO-if9to]

​@@ShauryaJain-kz1lk4×3×2×2

[@aaditjain5051]

Your answer at 1:15:04 is wrong Ans 1 is 34650 Ans 2 is 840 Final answer is 33810

[@1deepakx]

@@aaditjain5051pls check 🙏 what I have written 📝 🙂

[@AMANSEP]

Like if u have exam tomorrow 😂😂😂

[@urvashi-lw8nc]

How was the exam? Did this oneshot worked? ( I have my exam tomorrow)

[@gyangaming6295]

Yes brother I want it ​@@AMANSEP

[@sanatankagyan-fe5jk]

Bro i have 😂

[@btrgaming5965]

Compartment ka hai

[@PraveenSahai-l3r]

​@@btrgaming5965😂 same

[@Sumant-007]

16:20 I was ready for the concept And ushank sir is like swag he bhai hamara❤😂

[@subhamkumar1114]

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.

[@khushaboohariom9783]

😊

[@MahirajSinghRathore-vf8vg]

vedantu se copied lol

[@AnushreeUDAAN2024]

😂​@@MahirajSinghRathore-vf8vg

[@Kamransmgp0]

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".

[@admiralscott4119]

not 3 "I" its only 2

[@Kamransmgp0]

@@admiralscott4119 ok thanks

[@wkanadamz6wp]

Galat hai lod* answer 907200 hai 🍆

[@ashutoshupadhyay6629]

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 ❤❤❤😊😊😊

[@musicmusic2119]

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

[@--CHaHaL---]

YES JANN

[@radhikakhanna3300]

ending aate hue saare difficult aur important wale questions toh aapne home work de diye sir

[@Lopa617]

1:12:36 MONDAY - all vowels occurred together - 240 ways

[@SONALjain-m6s]

exactly!!

[@ArjunSharma-dq8ex]

54:12 Normal Answer =120 , even answer=48

[@sandeepjoyking]

How?

[@Benzene42]

5*4*3*2*1=120

[@Benzene42]

4*3*2*2=48

[@DishankuBharadwaj]

THIS LESSON COVERED ALL THE TOPICS OF NCERT ,,, THANK U SIR FOR THIS LECTURE , HOPE YOU WILL BRING SOME IMPORTANT QUESTIONS OF THIS CHAPTER

[@BristiSamanta-e7x]

10:48 30 ways 54:00 khud se nhi kar pai par normal 120 ways and even 48 ways 57:04 6 words with or with out meaning

[@jalajbansal4358]

39:25 Permutations means arrangement of something like A word,letter,number,etc

[@aakashvijay9538]

Hnn you are right

[@hardikkumar256]

Thank u sir ,aaj hi pura lecture dekh liya or samajh liya hai ......aiase hi padhate jaoo.... abhi extra question laga raha hu ...

[@utkarshshukla2219]

Aree hardik Bhai tussi idhar ke karr rahe ho 😂

[@hardikkumar256]

@@utkarshshukla2219 bus Bhai aapki cripa hai 😂😂

[@rishavranjan9429]

Me bhi aa gya 😎😎

[@Pranjwalkumar]

Swagat nahi karoge humara!?

[@hardikkumar256]

@@Pranjwalkumar aare kyo nahi Bhai

[@fogggamerz2316]

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

[@AhanaRoy-gn4ip]

54:11 ans 5p4 = 5!/(5-4) = 5×4×3×2×1=120 4p3= 4!/(4-3)= 4×3×2×1=24

[@reshmapathan8593]

Permutation=arrangement is important Combination=arrangement is not important

[@vijjiworld]

Thanks bro

[@reshmapathan8593]

@@vijjiworld My pleasure 😊

[@vijjiworld]

@@reshmapathan8593 bro now you are in which class

[@VISHAL_SAHARAN2007]

45:53 only jaat❤❤❤

[@pranavkumarkushwaha1740]

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!) .

[@justbehappyyyyyy]

1:51:05 Me , Akul and Lavi are those 3 students❤️🔥💕

[@endymion678]

Huhhh???

[@ShreyaKumari-b9j]

1:16:46 Total no of ways- 34650 Ans - 33180

[@ShanuTripathi-i2u]

33810 hoga

[@Scooby-Doo-l7u]

Legends are watching this just 2 hours before half yearly exams

[@kabiragarwal1365]

Us 😅😂

[@Scooby-Doo-l7u]

😂

[@Miyuki-OtakuSensei]

Me

[@arnavsharmaaaa]

Us ​@@Miyuki-OtakuSensei

[@17.txnzeem]

Bhai yeh one shot se kuch fayda hua?

[@Maggugaming]

1:47:21 Yadi EXAMINATION word ke Saare Letters ek dictionary me arrange kiye jaaye toh es arrangement meh E waale words start hone se pehle kitne words honge😊 TIP* : For This Type Of English Contact Me In IIT DHOLAKPUR😮

[@scenarionotes8757]

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

[@ADITYA-zg5qm]

Bhai same pov par mujhe pata h ki kisi teacher ko bolunga to vo Bina kisi explanation ke mujhe galat bol denge

[@OfficialManjeet2.O]

Kis kis ka exam 19 oct ko hai 😂

[@educationcentre6815]

Mera aaj hai

[@VinayakGupta427]

Mera

[@Ns_G.000]

Bhai tera kounsa school hai

[@fakesharma-d9z]

us bhai

[@Ibn_Rizwan]

Delhi Government 🙂

[@_Crafter_Manya_]

Best lect of PnC on whole KZbin fr❤👍 thanks a lottt Sir✨️

[@Maggugaming]

54:09 No. Of 4-Digit no. = 120 No. Of 4-Digit Even No. = 48

[@ganeshshinde08]

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!

[@santoshkumarsahu3905]

Sorry to correct you but your answer is wrong in the place of 9! It would be 10!

[@santoshkumarsahu3905]

Its also verified in ncert .

[@ananyasingh3448]

54:14 i)5!=120 and ii) 2P1 X 4P3 = 2! X 4! = 48

[@Artwithsakshi03]

Best lecture for one day before exam😊

[@visheshsharma4091]

Thank you sir this chapter is very important for ours

[@NXTJODCOBRA]

Best teachers for maths for class 11th on youtube❤😊

[@ravanxgaming4565]

54:13 , 120 total no. Of ways and 48 even no. Of ways

[@aryangupta378]

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.

[@PreetJain-m9b]

Sir mid term ki tayaari shuru karva do

[@JustForStudy_]

54:15 Total mein 120 but even wale mein 48 sahi hai kya??

[@shreyaarya5422]

Thank u sir jarurat thi iski 😢.....aur bhi chapter thoda jaldi laiye sir pls 🙏

[@mukulsharma4409]

Acha 🙁 sachi batti Or hamro jarurt na h kya 😂

[@krishivlalmazumder30]

@@mukulsharma4409 aukat mein reh... sir tera nokar nhi hein ... tu bachi hein aukat mein rahiyo

[@f__r__k]

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

[@Nighttalks_164]

Bhai samjh nahi ara

[@shreyanshgoel9158]

Sahi hai bhai

[@PushpaDevi-vj7qx]

1:48:23 Rank of word in the dictionary 😊😊 Thank you sir for the great lecture ...😊

[@ShobhaMishra-j9c]

Anyone from 2024-25😊✅

[@YesJee-m4j]

Thanku so much sir ❤ .......for this video ..😊 mai ak JEE aspirant hu or kbse permutations and combinations ka best lecture dhund rha tha pr koi smjh me he nhi aa rha tha ... Aapke wala lecture is OP ❤ ... sidha dimag ke raaste gya hai 😅.. thanku sir ❤

[@aanya_2013-v]

Sir actually sequence and series is also coming in mid terms so please bring a lecture on it too......😢

[@vijjiworld]

No bro

[@PhoenixX6T4]

1:40:49 Daughter wale question mai permutation use hona chahiya tha quki block bhi apna under rearrange hosakta hai

[@shamanchandra7414]

Like if your exam is today and your watching 3hours before😂

[@radhamishra2644]

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

[@mdbabafacts7790]

Live class vote for class 11👉👉👉👉👉

[@RatneshGaurawSrivastava]

Live class

[@JayShaIPL]

54:15 120 Even-48 Sir heart dedo jldi abb fir 1 ghari baat bolunga😂❤

[@JayShaIPL]

@@Krishna_g09200 waah waah bete!!🤩🤩

[@JayShaIPL]

@@Krishna_g09200 Tum bolo baccha 🌚

[@sciencewithtwist5770]

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.......

[@Riya-d6w8r]

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

[@alvinkumar96]

I completed the same in 45 minutes...

[@MayankSinghKapkoti]

1:37:53 Dammnnnn bro.Not gonna lie man my cycle used to have same lock no. 💀💀💀😳😳

[@Disha-w8s]

😂😂

[@kingb7]

54:07 no.of ways=120 FOR EVEN NO . = 48

[@Nancybarsaiyan]

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❤

[@DevanshMitta]

not except E but letter before E in alphabatic order (ABCD)

[@jatinthakur7168]

❤

[@OMNIVERSE0001]

1:56:58 SIR WE WANT LIVE CLASS 🙏🙏 FOR CLASS 11 MATHS CHAPTERS. 🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥

[@SHIVAYGupta-l9j]

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

[@zonecobragameplays6767]

Fake

[@sudhanshutomar7590]

Faiku

[@bxreco_edit]

Abe pheku ye bas ncert level ke questions hai

[@Political_dadaji]

Matlab kuch bhi

[@brosbeforehoes437]

Kitne paise liye?

[@rahimmiya1450]

1:47:25 question me kaha hai sare words ko ek dictionary me list kiye gaya hai to E alphabet se shuru hone wale word se pehle kitne words aate hai wo batana hai. Am I Right sir

[@Sayeed675]

0:00 Indroduction 2:50 Fundamental principles of Counting 27:02 Factorials 39:09 Permutation 1:22:47 Combination

[@kuldeepsolanki1767]

Exceptional ,short of words so ....impressive

[@farhankhan__588]

5:55 bro got emotional 😂

[@arunap6772]

11:17 Girls can get ready in 82,800 wayss

[@DragonATnight]

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

[@elitecutipiegamer2005]

11:11 82,800 ways that girls are ready for date 🌹😂😂😂😂baap re.....😂😂❤🎉

[@migueldiazcobrakai944]

54:16 Normal -----> 120 [5!] Even -----> 48 [4! × 2!] Where is my heart ❤️ 😑 ?

[@anilguptta1908]

in your body

[@-ar_yan]

bro this video is 11 months old

[@migueldiazcobrakai944]

@@anilguptta1908 yt heart brother 💀😭

[@migueldiazcobrakai944]

@@-ar_yan well I know that 🥹

[@shashikumardwivedi-p9g]

Becoz of u sir i got full marks in maths class11 😊

[@Aman-s5w]

Me ne bhi

[@Omprakash-t6n2n]

Today is my birthday and math exam😢.... 19 oct 2024...5.02 am....❤

[@Abhaysharma-i7f]

sorry bro late ho gya video dakna ma happy birthday

[@nugget8194]

25:53 sir isme 5×3 hona chaiye na ? Kyunki 2 flags toh use hogya .

[@AiusHR]

Sir plz mid term ke liyeee marathons upload kr doo😭😭😭😭😭😭😭😭😭😭😭😭

[@Lakshix]

Tere mkc

[@neelamdev-gu9mg]

1:48:18 ssirr examination me diiffferrent permutation ke word joo e se pahle aaayege so words starting with ABC and D

[@Vasudhm_kutumbkm]

Only A

[@Avni-k4e]

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

[@mayankdhami9956]

🧠🧠

[@TSHUKISHIRO]

A-an p-padh anpadh

[@bhavnimadaan998]

Btw pointing out ushank sir is so sweet he's calling even kiaan his own bacha even tho he's Ashu sirs son such an amazing teacher and such as amazing person

[@Painff-69]

bhai padle 😍

[@RaunakrajSingh-b7t]

35:15 sir roasted whole anti-national community just in few words 🗿🗿

[@TSHUKISHIRO]

Padhai me man laga😡

[@I_am_doomed]

​@@TSHUKISHIROWahi toh Faaltu ki debate me Lage hai sab. India=Bharat. Padhai chhodkar baaki sab karna hai

[@simplestudy31]

54:15. Normal answer-120 Even answer-48

[@kartik11pathak]

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!!

[@dishamathur8088]

But I is repeating so the letters starting with A should be 10!/2!

[@Aarnyx]

I is also repeating

[@MohitSingh-or4mh]

@@dishamathur8088 Exactly 👍

[@ujjwalkumar4523]

907200 is the answer 😊

[@georgeisaac1464]

​@@dishamathur8088 Bhai n is also repeating

[@_Soumya_0781]

Plz sir difficult question kariye 😢😢 49:25

[@uk_lofi]

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.

[@NotAkshat1069]

54:14 Normal - 120 Even - 48

[@migueldiazcobrakai944]

👍

[@TheycallmeA9bhav]

1:15:03 1663200/4×3×2×2 = 4989600

[@ShreyaKumari-b9j]

Thnx I'm finding the for few minutes finally I found it

[@chitaradhruv5663]

​@@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 😅

[@Radhe_radhe1201]

Like if u are studying just before the exam day for revision 😂😂

[@Ayushsharma001-q2i]

1:41:48 sir yeh 5 factorial ka concept samajh nhi aya.

[@tntumarp5614]

Sir never hesitate 😂😂

[@tarunsharma9278]

1:16:41 33810=permutation of mississippi I do not coming togather

[@KenXplains]

Fantastic lecture 👍

[@TSHUKISHIRO]

Thanks❤

[@MihirdeepKhavad]

11:10 82,800 ways

[@arunvishwakarma9266]

Sir iske baad sequence and series ka Lana please

[@Benzene42]

Gendu

[@harshilkakani]

1:47:35 the meaning is just simple that in how many ways can we permute the letters before E so just say that we have E letter is fixed hene no. of ways in which we can permute is 10! total ways= 10!/no. of repetitions 10!/2!*2!