السلام علیکم جناب ان ویڈیوز کو ہمارے لیے اپ لوڈ کرنے کے لیے آپ کا بہت بہت شکریہ،،،،،، جناب براہ کرم باب 7 تمام مشقوں کو اپ لوڈ کریں۔ جتنی جلدی ممکن ہو کیونکہ ہمارے پیپرز دسمبر کے پہلے ہفتے میں شروع ہوں گے جس میں پہلے 7 باب شامل ہیں
Sir i cannot explain how helpful these lectures are in 9th 10th i got not more than 40% ever in mathematics except boards but now i have totally understood each and every concept all because of the way you teach mathematics which is better than my subject teacher JazakAllah Sir Thanks Alot
Asslamoalikum Sir, The Answer you gave of Question#18 is a bit incorrect as you did not subtract the arrangements in which the zero comes in first place,ie in the total arrangements of 4 digit numbers some have 0 at the 1st place and the number automatically becomes a 3 digit number which has already been included in the 3 digit numbers category. So we will have to suntract those arrangements by fixing 0 at first place. The correct method and the answer is given below. Let's carefully solve the entire problem step by step. The goal is to find how many odd numbers less than 10,000 can be formed using the digits {0,2,3,5,6} without repetition. Key Observations: Odd numbers must end in 3 or 5. Numbers are divided into 1-digit, 2-digit, 3-digit, and 4-digit cases. Special care is needed when 0 is used, as it cannot appear in the leading position of any number. Case 1: 1-digit numbers The only odd digits are 3 and 5. So, the total count is: 2 numbers: 3,5 Case 2: 2-digit numbers The last digit (odd) can be 3 or 5 (22 choices). The first digit can be any of the remaining 4 digits (0,2,3,5,6 minus the chosen last digit). Without restrictions, this would give: 2×4=8 numbers. However, if the first digit is 0, the number becomes a single-digit number (e.g., 03, 05), which is already counted in Case 1. These invalid numbers are: 03 and 05 (2 invalid cases). So, the total for this case is: 8−2=6 numbers. Case 3: 3-digit numbers The last digit (odd) can be 3 or 5(2 choices). The first digit cannot be 0 (to ensure it's a 3-digit number), so there are 3 valid choices (2,6, and one other digit excluding the last digit). The middle digit can be any of the remaining 3 digits (after excluding the first and last digits). Thus, the total for this case is: 2×3×3=18 numbers. Case 4: 4-digit numbers The last digit (odd) can be 3 or 5 (2 choices). The first digit cannot be 00 (to ensure it's a 4-digit number), so there are 3 valid choices (2,6, and one other digit excluding the last digit). The second digit can be any of the remaining 3 digits (after excluding the first and last digits). The third digit can be any of the remaining 2 digits (after excluding the first, second, and last digits). Thus, the total for this case is: 2×3×3×2=36 numbers. Total Numbers Adding all cases: 2 (Case 1)+6 (Case 2)+18 (Case 3)+36 (Case 4)=62. Final Answer: The total number of odd numbers less than 10,000 that can be formed using {0,2,3,5,6} without repetition is 62.
1:53:08 sir your method is also correct but I think that we are allowed to place a consonant on third place because if we are not allowed to place it then it is impossible to do even a single permutation And also the word "may" is used in the question ,which shows the allowance of placing a consonant on the third place If I am wrong then sir please kindly correct me
Sir in some questions you are adding total arrangements and in some you are multiplying permutations to get end result, how should we know when to multiply and when to add???
thank you soo much sir app ki nahe pta moje is video se bahooooot help mily mei jb exercise dekhi to mei ny kaha ye baoot tough ho gi ...but thank you ye mery leye asan ho gae app ki waja seee
Sir q12 ghalat hai. Let me explain. Yahan contrary method kaam nahi karega. The reason is that the contrary method only works if something can happen or something cannot happen at all, meaning there is no third option. Contraty method only works if there are only 2 possible choices. Contrary method doesnt work here becuse there are 3 possibilities. The vowels may occupy only even or only odd or they occupy both odd and even, thats why contrary method doesn't work. Instead, this is how it should be done: Given: Given Word 'VOWEL' = 5 letters No. Of vowels = 2 No. Of consonants = 3 No. Of odd places = 3 Permutations of the vowels occupying odd places = 3P2 = 6 (Basically we are choosing 2 spaces from those 3 to put out vowels in) Remaining places = 5-2 = 3 (we put 2 vowels in the odd places so 3 spaces remain from 5) Permutation of consonant letters in the remaining place = 3! = 6 So, the permutation of the word 'VOWEL' in which vowels may occupy odd places = 6×6 = 36 Isko pin kardijiye taa keh durso ko pata ho aur exam mai ye ghalti na ho, shukaiya!
Sir aik confusion hai k jin questions mai akhir mai total arrangements find krni hain un mai kuch mai add kia hai kuch mai multipy,ye kesy pata chalta hai,aik clear nishani?
@@raplineaddict I actually forgot since this was a bit long ago . But I assume that you have the same doubt as me that why do we add permutations in some questions and add in other questions . Right ? Okay so the logic is that when various permutations can happen simultaneously, we multiply them, but when there are diff cases like for odd and even wale questions , we add them , cause the number will either have 3 ,5 7 ,9 or 1 at a time . There can only be one of these at a time , so in this case we'll add the permutations obtained for diff cases . Hope it helps !
@@nimrahussain5710 thanks a lot buddy! it really helped. to add to this, even if we multiply in cases walay questions, like multiplying the permutation with the number of cases, we'll get the same answer.
Sir, ref Ex. 6.2 Q. 13, confusion in solving... You considered repetition of vowels in 1st, 3xposition of the word, but repetition may occur in other 3x positions, i.e., 2nd, 3rd, and 4th, etc. As per my calculation the ans is 1290, because there 5 x possible Positions of repetition and in each possible arrangements are 720, so total is 3750 nos of repetition & total nos of possible arrangements are 5040, so without repetition arrangements are 5040-3750=1290 Nos.
2:08:00 Sir I am having a confusion in this question ,as it is not mentioned in the question to use all the digits at the same time ,so can't we also count numbers having 2-digits and similarly 3,4,5 digit numbers can be counted I will be thankful to you for doing my correction
Sir Question 6 may aap ne baat ki k agar number repeat hon gy tu sari places pr wohi number aay ga jesay aap ne 6 put kiya. Sir agar hum 5 put karain ya 4 put karain tab tu answer different aay ga ye smjha dein zra
Sir apne kaha k question12 and 13 jese questtion ko teacher according to book krlety jo k ghalat ha ...r hm sahi kr rhyy lkn board exams to unho ne hi check krmy agr wo ghalat krdee hmryy answer then????what should we do now Should we follow books or your lectures
Dear don't worry I will explain question 4 m even pochy ty os ny to 2,4,6 ya 3 even hn es ly 3 case banay ab question 5 wahan py os ny just kaha k first py 5 Baki Jo marzi to per 1 hi case banay ga na or 6 to simple ha k reputation allow to wo simple ho solve ho jay ha
and sir in q16 why are we not considering that these numbers can also form 2 3 4 digit odd numbers as well , why are only considering 5 digit odd numbera
Aoa sir, sir in question 13, the total no of words are 7 and vowels are fixed which means 3 words are fixed so how 5 other words are left isn’t it supposed to be left with 4 words?
To find the number of permutations, we need to calculate the factorial of the number of letters. The word "SHUMAIL" has 7 letters. 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040 So, there are 5040 different permutations that can be made from the letters in the word "SHUMAIL".
Permutation wale questions me order matter kra hay or ese words use hote hain"arrange,line up ,rank" jab ke combination me order matter nhi krta or ese words use hote hain "choose ,select ,pick ,group" is se app ko pata la jae ga pperme ke konsa question permutation ka hay or konsa combination
Sir last question 23 waly ki kuch sense ni bni samjh no aya ky wo MULTAN kesy bana????? I mean agra given word ky according (1,2,3,4,5,6) ki permutation krein tou MULTAN ni banta aur agar given permutation(4 6 3 2 1 5) ky according bhi krein tou MULTAN ni banta ........ Please iss ko dubara explain kr dein ap..
@@Calculus.CornerActually in 3rd step you are taking 2 common from (2n-2) but I am not taking common. I am just multiplying (2n-1)(2n-2) Taking common makes the simplification easier, it does not really affect the ans, right?
Sir how will we identify ke ye question permutation ka ha ya phir combination ka because questions are very similar?
permutation mein repeate likha hoga
Permutation basically arrangement hota hai aur combination main selection hoti hai yeh main difference hai.
Hope you understand
muqadas
total words: 7
repeated: A(2) times
permutation:7!/2!
2520 permutations
Exactly 💯
Good
السلام علیکم جناب ان ویڈیوز کو ہمارے لیے اپ لوڈ کرنے کے لیے آپ کا بہت بہت شکریہ،،،،،، جناب براہ کرم باب 7 تمام مشقوں کو اپ لوڈ کریں۔ جتنی جلدی ممکن ہو کیونکہ ہمارے پیپرز دسمبر کے پہلے ہفتے میں شروع ہوں گے جس میں پہلے 7 باب شامل ہیں
G InshaAllah
Inti halish Urdu wah
Thankyou so much sir!!
Congratulations on successfully crossing 10k subscribers! Many more milestones to come INSHA'ALLAH!✨🫶🏻
Thanks so much dear 💐💐
Sir you are very helpful (especially in exams 😂)... Stay blessed❤
Sahi baat ha shamshad
Maths is quite a difficult subject , your way of teaching makes it very intresting
thanks sir
Welcome dear 💐
Kon kon sendups ki tiyari kany aga hai?
Me😂
Me
usss
Mere
Resend up 🤧
very welll sir
you are doing great
We are grateful to Allah who gave us teachers like you
ALHAMDULLILAH
Thanks so much dear for your compliments 💐
Your way of teaching is very effective mashaAllah
Thanks for compliments 💐
sir bohat ajeeb si excwercise thi lekin samhj aa gai app ki wajah se shukria sir
Really appreciate your work sir. . Thanks 👍
Thanks and welcome
Sir i cannot explain how helpful these lectures are in 9th 10th i got not more than 40% ever in mathematics except boards but now i have totally understood each and every concept all because of the way you teach mathematics which is better than my subject teacher JazakAllah Sir Thanks Alot
Welcome Dear 💐
Always remember guys if your answer comes in decimal your answer is incorrect
Asslamoalikum Sir,
The Answer you gave of Question#18 is a bit incorrect as you did not subtract the arrangements in which the zero comes in first place,ie in the total arrangements of 4 digit numbers some have 0 at the 1st place and the number automatically becomes a 3 digit number which has already been included in the 3 digit numbers category. So we will have to suntract those arrangements by fixing 0 at first place.
The correct method and the answer is given below.
Let's carefully solve the entire problem step by step. The goal is to find how many odd numbers less than 10,000 can be formed using the digits {0,2,3,5,6} without repetition.
Key Observations:
Odd numbers must end in 3 or 5.
Numbers are divided into 1-digit, 2-digit, 3-digit, and 4-digit cases.
Special care is needed when 0 is used, as it cannot appear in the leading position of any number.
Case 1: 1-digit numbers
The only odd digits are 3 and 5. So, the total count is:
2 numbers: 3,5
Case 2: 2-digit numbers
The last digit (odd) can be 3 or 5 (22 choices).
The first digit can be any of the remaining 4 digits (0,2,3,5,6 minus the chosen last digit).
Without restrictions, this would give:
2×4=8 numbers.
However, if the first digit is 0, the number becomes a single-digit number (e.g., 03, 05), which is already counted in Case 1. These invalid numbers are:
03 and 05 (2 invalid cases).
So, the total for this case is:
8−2=6 numbers.
Case 3: 3-digit numbers
The last digit (odd) can be 3 or 5(2 choices).
The first digit cannot be 0 (to ensure it's a 3-digit number), so there are 3 valid choices (2,6, and one other digit excluding the last digit).
The middle digit can be any of the remaining 3 digits (after excluding the first and last digits).
Thus, the total for this case is:
2×3×3=18 numbers.
Case 4: 4-digit numbers
The last digit (odd) can be 3 or 5 (2 choices).
The first digit cannot be 00 (to ensure it's a 4-digit number), so there are 3 valid choices (2,6, and one other digit excluding the last digit).
The second digit can be any of the remaining 3 digits (after excluding the first and last digits).
The third digit can be any of the remaining 2 digits (after excluding the first, second, and last digits).
Thus, the total for this case is:
2×3×3×2=36 numbers.
Total Numbers
Adding all cases:
2 (Case 1)+6 (Case 2)+18 (Case 3)+36 (Case 4)=62.
Final Answer:
The total number of odd numbers less than 10,000 that can be formed using {0,2,3,5,6} without repetition is 62.
Exactly my apsay totally agree krta mera be yahi ans tha.
Sir you should be education minister of Pakistan 🇵🇰 ,our country needs teachers like you ❤
Thanks dear
Sir keep it up.
Many more things are coming to you.
❤
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1:53:08 sir your method is also correct but I think that we are allowed to place a consonant on third place because if we are not allowed to place it then it is impossible to do even a single permutation
And also the word "may" is used in the question ,which shows the allowance of placing a consonant on the third place
If I am wrong then sir please kindly correct me
Yes you are right 👍
sir dil sey dua nikalti ha apk liyeh boht acha parha rhy han
Thanks so much dear 💐
TAYYAB
Total number of words =6
Total number of choices=6
Repeated A(2) and Y(2)
6 factorial÷2 factorial * 2 factorial
= 720÷4
=180
Good 👍
Sir, your lectures are excellent, and your way of explaining is so amazing that we can't even imagine. May Allah grant you a long life. 🙏😊
Thank you so very very much...🎉🎉🎉your lectures are really helpful for us 🎉🎉🎉🎉🎉.May Allah bless you with many more happiness 🎉🎉. Aaemen 🤲
Thanks and welcome
Sir agr kisi question mein n ki both values negative ajaye tu question is not possible?
Yes it's mean question m koi mistake ha
From Atif Ahmad Official ❤
💐
Phy wle?
Yes
HUZAIFA
Total numbers= 7=n
No. of choices=7
Letters repeated=A(2)=n1
Permutations=n!/n1!
So,
7!/2!=5040/2
2520
Good 👍
Sir in some questions you are adding total arrangements and in some you are multiplying permutations to get end result, how should we know when to multiply and when to add???
Dear explain in vedio lectures
thank you soo much sir app ki nahe pta moje is video se bahooooot help mily mei jb exercise dekhi to mei ny kaha ye baoot tough ho gi ...but thank you ye mery leye asan ho gae app ki waja seee
Welcome Dear
Sir q12 ghalat hai. Let me explain.
Yahan contrary method kaam nahi karega. The reason is that the contrary method only works if something can happen or something cannot happen at all, meaning there is no third option. Contraty method only works if there are only 2 possible choices. Contrary method doesnt work here becuse there are 3 possibilities. The vowels may occupy only even or only odd or they occupy both odd and even, thats why contrary method doesn't work.
Instead, this is how it should be done:
Given:
Given Word 'VOWEL' = 5 letters
No. Of vowels = 2
No. Of consonants = 3
No. Of odd places = 3
Permutations of the vowels occupying odd places = 3P2 = 6
(Basically we are choosing 2 spaces from those 3 to put out vowels in)
Remaining places = 5-2 = 3 (we put 2 vowels in the odd places so 3 spaces remain from 5)
Permutation of consonant letters in the remaining place = 3! = 6
So, the permutation of the word 'VOWEL' in which vowels may occupy odd places = 6×6 = 36
Isko pin kardijiye taa keh durso ko pata ho aur exam mai ye ghalti na ho, shukaiya!
Right 👍
Thankyou so much sir ... Your method of teaching is awesome🥰🥰
Most welcome dear 💐
2:19:57 sir there's mention without repetition of digits but you have repeated zero in 5 Digit case.
A huge respect for you 🎉⭐⭐⭐
Welcome Dear 💐
Very Informative lecture Sir...❤
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Sir aik confusion hai k jin questions mai akhir mai total arrangements find krni hain un mai kuch mai add kia hai kuch mai multipy,ye kesy pata chalta hai,aik clear nishani?
May Allah give u jaza e kair❤
SIR YOUR TEACHING METHOD IS SOOO GOOD!!
Thanks and welcome
sir my question is that we multiplied permutations in all questions but added them in 2 question , why
Explain in vedio lectures please listen carefully again
yaar ye mujhay bhi samjha do
@@raplineaddict I actually forgot since this was a bit long ago . But I assume that you have the same doubt as me that why do we add permutations in some questions and add in other questions . Right ? Okay so the logic is that when various permutations can happen simultaneously, we multiply them, but when there are diff cases like for odd and even wale questions , we add them , cause the number will either have 3 ,5 7 ,9 or 1 at a time . There can only be one of these at a time , so in this case we'll add the permutations obtained for diff cases . Hope it helps !
@@nimrahussain5710 thanks a lot buddy! it really helped. to add to this, even if we multiply in cases walay questions, like multiplying the permutation with the number of cases, we'll get the same answer.
Amazing..style of teaching
Thanks 💐
In college di you teach?
Federal Government
You are a good teacher
sir you deserve million of subscribers
Thanks Dear 💐
Thank you sir for guidance.Sir maths k liay aap konsi guide prefer krty hain ?
No guide only text book read krain or practice jitni zeada krain gy atny marks zeada ain gy
Sir, ref Ex. 6.2 Q. 13, confusion in solving... You considered repetition of vowels in 1st, 3xposition of the word, but repetition may occur in other 3x positions, i.e., 2nd, 3rd, and 4th, etc. As per my calculation the ans is 1290, because there 5 x possible Positions of repetition and in each possible arrangements are 720, so total is 3750 nos of repetition & total nos of possible arrangements are 5040, so without repetition arrangements are 5040-3750=1290 Nos.
?
No dear your concepts is wrong please listen and this lecturer then may b you understand this concept and it's right answer and method 100%
2:08:00
Sir I am having a confusion in this question ,as it is not mentioned in the question to use all the digits at the same time ,so can't we also count numbers having 2-digits and similarly 3,4,5 digit numbers can be counted
I will be thankful to you for doing my correction
Jo time mention kiya ap ny os question m to odd number hn to add number ki condition hoti h k last py odd digit hota ha so esy hi solve kia ha
Sir you are legendary teacher of maths❤
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Sir Question 6 may aap ne baat ki k agar number repeat hon gy tu sari places pr wohi number aay ga jesay aap ne 6 put kiya. Sir agar hum 5 put karain ya 4 put karain tab tu answer different aay ga ye smjha dein zra
Thanks very helpful ❤
Welcome 💐
35:44 Bari value ki respect karni chahye
Sir mehtab being funny 😂❤
😀Kisi Tara to math samj ah jay
@Calculus.Corner 💗💞 bilkul sir g
U made maths easier for me
Sir apne kaha k question12 and 13 jese questtion ko teacher according to book krlety jo k ghalat ha ...r hm sahi kr rhyy lkn board exams to unho ne hi check krmy agr wo ghalat krdee hmryy answer then????what should we do now
Should we follow books or your lectures
Dear peli bt board m question wrong ni ata so don't worry
Sir question no 4 ma 3 digits thy to isi liye 3 case banay thy?
Yes
@@Calculus.Corner Sir iss tarha question 5 ma 7 digits or 6 ma 4 digits thy to wo to ek ek case ma hi solve ho gay
Asa kiuu sirrr????😭
Dear don't worry I will explain question 4 m even pochy ty os ny to 2,4,6 ya 3 even hn es ly 3 case banay ab question 5 wahan py os ny just kaha k first py 5 Baki Jo marzi to per 1 hi case banay ga na or 6 to simple ha k reputation allow to wo simple ho solve ho jay ha
Hope you understand and first thing please listen carefully to the lecturer otherwise the same way you concepts can't clear
@@Calculus.Corner Sir Tysm for your explanation a gaya samj😊
Best of luck 😊
Thanks 💖
and sir in q16 why are we not considering that these numbers can also form 2 3 4 digit odd numbers as well , why are only considering 5 digit odd numbera
Given in question that's why
Sir bohot help mili .mara sendup ma .thx❤❤❤❤
Thanks sir may allah bless you
Welcome dear 💐
Sir kich jagahon pr total apas mai add horhe kuch mai multiply wo kyu?
Explain Kiya ha dear m ny video lecturer m
@@Calculus.Cornersirrrr kis jagah pr pls dubara bta den ppr hone wala haiii
Excellent lecture sir👍
Thanks and welcome
2:15:17 sir zero KO exclude karna tha, q ki wo 1 digit numbers me a jata hay.. example 03 = 3 wich is already counted
No dear zero Ak digit ha os ko exclude ni kr skty
@@Calculus.Corner sir Mene is Ka solution nikala hay٫ zero KO start me nai Rakh kar٫ ہمیں start me non-zero digit fix karna hoga
Sir g Shayan name ki 360 repeated permutations banein gi na?
Many congratulations on crossing 10k subs you deserve a 1M 🙂, SOON .
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Aoa sir, sir in question 13, the total no of words are 7 and vowels are fixed which means 3 words are fixed so how 5 other words are left isn’t it supposed to be left with 4 words?
Explain in vedio lectures ni detail please listen carefully
Asslam o Alikum.Sir in Q-18 of Ex.6.2 03 and 3(1-digit number) and 05 and 5 are same.
Plz explain
W.s what about 0 is it digit or not ???
Sir we want to make number of digits less than10,000.therfore 03 and 3 r repeated.
sir can u kindly tell me that why u multiplied 2 with 6 in the 12 question its very confusing.
Dear just for simplification
Sir chapter 7 kab tak lagaenge? It’s included in sendups
please sir upload it as soon as possible
Sendup sy pelay inshallah upload ho jay ga
No it's not included only till chapter 6 for us
Sir Q 15 ko agr repeated permutation k method sy krain to 2520 answer ata ha book wala q k is mn bhi to Q 9 ke trahan repitition ha
Okay
@Calculus.Corner sir mtlb repeated permutation ka formula lga skty hain Q15 mn ya nhi ?
Sir plz clear my confusion
To find the number of permutations, we need to calculate the factorial of the number of letters.
The word "SHUMAIL" has 7 letters.
7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040
So, there are 5040 different permutations that can be made from the letters in the word "SHUMAIL".
Bhai chat gpt krnay se acha comment hi na krta😂
@mr.beanshorts5694 bro sai keh raha ru
@@muhammadshumailafzal5700 hahaha mazak kar raha hu mene bhi chat gpt pe search kiya tu chat gpt ne mujhe bhi same bànaky diya tha😂
Sir very very thank to you, for you hard work for us. Sir ALLAH ap ko khush rakhay orr duniya-o-akhrat ma kamyabi ata karyn AMEEM❤❤
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Hasaan=36 permutation 😅
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AYESHA. 360 permutations
Good
Allah sir AP na hoty tou hm Kahan jaty 😢. Allah apko hamesha sehat wa kamiyabi....
Ameeen😊
2:19:03 Sir zero fix Kia to wo four digit Hoga Jo ki already ham count Kar chunky hen.
No hum ny zero ko start py rkna ha i.e 0123,0124,.....
@@Calculus.Corner it will be same is 123٫ 124
@@muhammadzahir7931 was thinking the same thank God someone asked, I wondered how did no one thought tha5
Sir can we solve qs 15 with the same method as we did in qs 9? In this way, we get same answer as book pls guide...
Congrats Sir for 10K . More to come💛
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Adeel
Total letters=n=5
Repetition of letter e=2
Permutation=5 factorial \2 factorial=60
vary intaligant buoy
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AHSAN
No of words=5
A=2
5!/2!
120/2
60 Permutations Answer
Sir in question 4 why you not use fundamental low of multiplication
Explain in vedio lectures ni detail dear
sir kuch sawalo me last me aap permutations ko add karte hai aur kabhi multiply kartee hai ye explain kardee
Dear ya according to given condition hota ha Jo k explain Kiya ha m ny vedio ma ab text m kisy explain kron
sir question 15 KO rotational permutation sa karsakta ha jasa question 9 kia 2 blue 3 yellow 4 green flags wla ka jaisa
Agr circle permutation ban Rai hn to kr skty hn
Sir please upload Chapter 7 as it's included in send up syllabus. Our send ups are starting from the first week of December.
InshaAllah dear Nov m hi complete ho jay ga don't worry
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Welcome 👍
Sirthanks🎉🎉
Sir at 31:43 i get 2 answers 13 and 1 because i make it quadratic equation so kindly plz tell me is it correct or not?
If answer are same then it is right otherwise wrong
@Calculus.Corner your answer is only 13 and mine are 13 and 1
Ap permutations k m n ki jaga 1 put kr k check krain agr sai ay to per theek otherwise ap ny mistake ki ha koi 1 ni ay ha only 2 hi ho ga
Ok
Sir chapter 7 +9 plizzzz ....... Apki bhut achi smjh aty hai
InshaAllah dear uploaded soon 💐
Permutation wale questions me order matter kra hay or ese words use hote hain"arrange,line up ,rank" jab ke combination me order matter nhi krta or ese words use hote hain "choose ,select ,pick ,group"
is se app ko pata la jae ga pperme ke konsa question permutation ka hay or konsa combination
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I m teacher as well but Sir G You are lagend ❤❤❤❤❤
Sir last question 23 waly ki kuch sense ni bni samjh no aya ky wo MULTAN kesy bana????? I mean agra given word ky according (1,2,3,4,5,6) ki permutation krein tou MULTAN ni banta aur agar given permutation(4 6 3 2 1 5) ky according bhi krein tou MULTAN ni banta ........
Please iss ko dubara explain kr dein ap..
Dear very easy ap again sunain per b ni samj ati to mujy Whatsapp krain
Very easy. 123456 ke mutabik multan hi bnta hai
Sir 10 question main second part main total number 6 hain or choise no 5 hain
SIR Thanks You Too Much ❤ May Allah Blessed You 💗🤝
Ameen welcome dear 💐
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@@ArfaAbrar-b7x genuineeeeee?
Very nice Sir I Have ever seen 😍😇
Welcome dear
Sir hamarii teacher ap ki videos dakh ker school ati hayy😅😅
Same bro 😂
I think there is mistake in q 18
Bcz when u take in four digits permutation with 4 p 3
0 is also included
0453
No dear you think wrong
Yes I agree with you
A.O.A
I think question no 18 MN ap NY mistake ki Hy
Please check krain
Which step ??
@Calculus.Corner 03 and 05 are one digit number
Not two
Dear it's 2digit kia 0 ko ap digit ni consider krty ???
Similarly 025 and 023 are two digit numbers not three digit
@@Calculus.Corner is there any difference between 3 and 03 ?
Thank you so much respected sir❤
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kiya baat ha sir itni jaldi samaj a gai🙂
Chalo ab 2 din baad sare question khud solve ker ke check kerna 😂
@@mka5242 Sirf samaj ai ha 😁
Sir in Q 2 part ii, I am getting 2 +ve values for n
n=13 and n=1
Can you plz explain why???
Dear ap ny koi mistake ki ho gi otherwise esa ni hota
@@Calculus.CornerActually in 3rd step you are taking 2 common from (2n-2) but I am not taking common. I am just multiplying (2n-1)(2n-2)
Taking common makes the simplification easier, it does not really affect the ans, right?
Sir please 9th chp upload krden
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InshaAllah dear uploaded soon
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A L I
Total no of words = 3
Choices = 3
³P3 = 6
So 6 is answer. Please check it
Sir Q6 mn 6 he q lia ha hm koi bhi number nhi ls skty ?
?
No dear ap question ko carefully again listen krain samj ah jay gi k 6 hi q ay ga
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