Answer: 105 ?!?! x>=1 y >=3 z >=4 Then minimum sum x + y + z = (1+3+4) =8 So we require ( 21 - 8) = 13 coin to distribute in 3 beggar(x,y,z) (13+3-1)C(3-1) = 15C2 = 105
Since a dice’s outcome can’t be less than 1 So in order to remove the cases of (0,-1,-2) First we give 6 (-1) coins to one beggar and even after giving these 6 coins we still include the beggar in distribution, so that it may get total 7 coins or 8 coins
Since a dice’s outcome can’t be less than 1 So in order to remove the cases of (0,-1,-2) First we give 6 (-1) coins to one beggar and even after giving these 6 coins we still include the beggar in distribution, so that it may get total 7 coins or 8 coins
@factorialacademy Sir please make a video on how to understand these things how to remove cases like this abhi bhi poora nahi samajh aya 😓😓 Kya hamesha same cheez hi karenge sirf ek ko disclude karenge kya aisa nahi ho do bhi disclude karna pade for example if the dice has outcomes only from (1-7)
sir I do not understand that why in both the questions we do not minus the excessive cases, like for example in 2nd que we subtracted the case where 6 coins are given to one beggar but not the case where 7 or 8 coins are given to one beggar
Since a dice’s outcome can’t be less than 1 So in order to remove the cases of (0,-1,-2) First we give 6 (-1) coins to one beggar and even after giving these 6 coins we still include the beggar in distribution, so that it may get total 7 coins or 8 coins
7:00 sir aise toh saare possible outcomes aayege na voh bhi toh ayega agar ek hi beggar ko saare coins de diye tab toh uske paas 6-8=-2 ho jayege aise -1 bhi agar 7 coins diye tph voh saare possible outcomes bhi toh remove karne hoge
Beggar ko 6 (-1) coins dene ke baad bhi uss beggar ko distribution me include kiya hai Toh ab use 0 (-1) coin 1 (-1) coin 2 (-2) coins yeh sare possible cases included hai method me
@@factorialacademy oh acha got it so like 7:26 par ek random ko 4C1 karke choose kara and usse 6(-1) ke coins diye and then firse 2 bache hue (-1) ke coins baant diye among them and iss puri possibility ko total mei se minus kar diya got it sir thank you 👍👍
Q type which include remianders numericals or qs which first make u try to solve using cases bcuz that's what's really required rn. Prolly before mains ,last date hai meri so i can wait till then. But rlly need numerical pnc solns
Since a dice’s outcome can’t be less than 1 So in order to remove the cases of (0,-1,-2) First we give 6 (-1) coins to one beggar and even after giving these 6 coins we still include the beggar in distribution, so that it may get total 7 coins or 8 coins
El particular beggar jise 4 coins diye, uske baad bhi us beggar ko distribution me include kiya hai Toh ab use ek aur coin extra milne wala case bhi amswer me included hai
Since a dice’s outcome can’t be less than 1 So in order to remove the cases of (0,-1,-2) First we give 6 (-1) coins to one beggar and even after giving these 6 coins we still include the beggar in distribution, so that it may get total 7 coins or 8 coins
So basically, in order to give 8 (-1) coins to one begger, you must first give him 6 (-1) coins. Then you have 2 (-1) coins left. in the video, rather than excluding the -6Coins Begger when distributing the remaining 2 (-1) coins, he includes him, therefore the cases where the begger with -6 coins can get another one or two (-1) coins is also included.
The maximum we can have in a box is 4, if while disturbing 5 coins to a beggar if we give 4 by accident then the primary condition is not sastified so that case would be extra... so we are subtracting it
@JadenIIT we can't give all the 5 ( ie already one is present and we give all the five remaining )to one beggar na , so shouldn't we subtract that case too...we can maximum give 3 cux one is already there and we have 5 so there's a chance we can give all 4 to one or all 5 to one ....so doesn't two cases arise
Please continue this series sir! Its fun to revise such small concepts during Dinner / Lunch
For the first problem
We can also evaluate coefficient of x^12 in
(1 + x + x^2....).(1+x^2+x^4+...).(1+x^3+x^6....).(1+x^4+x^8....)
Multinomial theorem waale approach se bhi pnc ke questions laaiye
Okay will put
both done by generating function ( not very ideal to use here. but just wanted to refresh this topic)
15C2 is the answer!
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Please sir keep posting PNC videos, much needed, your definite integral series was awesome it made my integration a lot stronger !!
Glad you liked the integral series! 😄
Will bring more PnC 👍
105! i was not able to solve such questions tysm this is an amazing way solved it within seconds
Great work, keep practicing these types of problems!
This approach is very nice earlier I was stuck on hint and trial😢
Glad to help.
Answer: 105 ?!?!
x>=1 y >=3 z >=4
Then minimum sum
x + y + z = (1+3+4) =8
So we require ( 21 - 8) = 13 coin to distribute in 3 beggar(x,y,z)
(13+3-1)C(3-1) = 15C2 = 105
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thank youu sir for considering the hw suggestion, it indeed helps a lot. The HW ans is 105
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Glad you found it helpful. 😊
Multinomial theorem for the second one ? 🤔 But still got something new to learn❤
seems like this method in gonna help me so much, even apart from academics-
most welcome
105 solved thank sir for such great content
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Welcome!
7:37 Sir isme hamlog wo cases kyun nhi minus kr rhe h jisme 1 beggar ko 7 (-1 ) coin or 8 (-1) coin mil gaya ho
Since a dice’s outcome can’t be less than 1
So in order to remove the cases of (0,-1,-2)
First we give 6 (-1) coins to one beggar and even after giving these 6 coins we still include the beggar in distribution, so that it may get total 7 coins or 8 coins
@@factorialacademy Thank you sir❤
7:32 Bache hue 2 (-1) ke coin 4 logon mein baatne hain ya 3 logo mein kyuki ek ko to 4 (-1) ke coin pehle hi de chuke
Since a dice’s outcome can’t be less than 1
So in order to remove the cases of (0,-1,-2)
First we give 6 (-1) coins to one beggar and even after giving these 6 coins we still include the beggar in distribution, so that it may get total 7 coins or 8 coins
@factorialacademy
Sir please make a video on how to understand these things how to remove cases like this abhi bhi poora nahi samajh aya 😓😓
Kya hamesha same cheez hi karenge sirf ek ko disclude karenge kya aisa nahi ho do bhi disclude karna pade for example if the dice has outcomes only from (1-7)
sir I do not understand that why in both the questions we do not minus the excessive cases, like for example in 2nd que we subtracted the case where 6 coins are given to one beggar but not the case where 7 or 8 coins are given to one beggar
Since a dice’s outcome can’t be less than 1
So in order to remove the cases of (0,-1,-2)
First we give 6 (-1) coins to one beggar and even after giving these 6 coins we still include the beggar in distribution, so that it may get total 7 coins or 8 coins
@@factorialacademy okay sir thank you very much got it.
need !!
more videos on PNC 3D vector
Okay will do👍
7:00 sir aise toh saare possible outcomes aayege na
voh bhi toh ayega agar ek hi beggar ko saare coins de diye tab toh uske paas 6-8=-2 ho jayege aise -1 bhi agar 7 coins diye
tph voh saare possible outcomes bhi toh remove karne hoge
Beggar ko 6 (-1) coins dene ke baad bhi uss beggar ko distribution me include kiya hai
Toh ab use 0 (-1) coin
1 (-1) coin
2 (-2) coins yeh sare possible cases included hai method me
6 already krdiye aur jo 7 aur 8 h vo dono bhi usme hi aa gye cuz jo 2 bche hue h unhe sbme distribute kra h toh 7 aur 8 dono uske hi cases h
@@factorialacademy oh acha got it
so like 7:26 par ek random ko 4C1 karke choose kara and usse 6(-1) ke coins diye and then firse 2 bache hue (-1) ke coins baant diye among them
and iss puri possibility ko total mei se minus kar diya
got it sir thank you 👍👍
Q type which include remianders numericals or qs which first make u try to solve using cases bcuz that's what's really required rn.
Prolly before mains ,last date hai meri so i can wait till then. But rlly need numerical pnc solns
7:42 yha apne minus kaise kra ye ? kyu kra vo toh smjh gya
Kyunki dice ka outcome 0 nhi ho skta
Since a dice’s outcome can’t be less than 1
So in order to remove the cases of (0,-1,-2)
First we give 6 (-1) coins to one beggar and even after giving these 6 coins we still include the beggar in distribution, so that it may get total 7 coins or 8 coins
Sir ye -ve coin method kaise soocha aur kya app 2nd question ko standard way mai explain kar sakthay hai kya ek baar
Pahle wale question me minus kyo kiya hai apne baad me ??
Ho chuka Hai means what?
‘It has been done ‘
@factorialacademy dhanyavaad
1st question me 7 aur kyu minus nhi kiye jab kisi ek ko pure 5 coin milenge
El particular beggar jise 4 coins diye, uske baad bhi us beggar ko distribution me include kiya hai
Toh ab use ek aur coin extra milne wala case bhi amswer me included hai
Homework ans is 15C2=105
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the case in second ques when we give -8 coins to single beggar i could not understand or is missing?
Since a dice’s outcome can’t be less than 1
So in order to remove the cases of (0,-1,-2)
First we give 6 (-1) coins to one beggar and even after giving these 6 coins we still include the beggar in distribution, so that it may get total 7 coins or 8 coins
So basically, in order to give 8 (-1) coins to one begger, you must first give him 6 (-1) coins. Then you have 2 (-1) coins left. in the video, rather than excluding the -6Coins Begger when distributing the remaining 2 (-1) coins, he includes him, therefore the cases where the begger with -6 coins can get another one or two (-1) coins is also included.
Hw ans 105?
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Yes right
Hw ans 105
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Hw question direct application of beggers method 105 answer
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Is the answer 105?
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Bhai ek video skew lines par bana do pls
Hw question ans 105
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105
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@factorialacademy can you make a video on important models of calculus?
I didn't understand the subtraction in first question 😭😭😭
The maximum we can have in a box is 4, if while disturbing 5 coins to a beggar if we give 4 by accident then the primary condition is not sastified so that case would be extra... so we are subtracting it
@JadenIIT we can't give all the 5 ( ie already one is present and we give all the five remaining )to one beggar na , so shouldn't we subtract that case too...we can maximum give 3 cux one is already there and we have 5 so there's a chance we can give all 4 to one or all 5 to one ....so doesn't two cases arise
@@DLokesh-lc4bq you take maximum case the lower cases are handled that is why beggars is applied
105
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105
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105
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