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
@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)
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
7:00 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 👍👍
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
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.
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
Multinomial theorem waale approach se bhi pnc ke questions laaiye
Okay will put
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 👍
Multinomial theorem for the second one ? 🤔 But still got something new to learn❤
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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This approach is very nice earlier I was stuck on hint and trial😢
Glad to help.
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. 😊
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 ye -ve coin method kaise soocha aur kya app 2nd question ko standard way mai explain kar sakthay hai kya ek baar
need !!
more videos on PNC 3D vector
Okay will do👍
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
Hw question direct application of beggers method 105 answer
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 👍👍
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
Pahle wale question me minus kyo kiya hai apne baad me ??
Hw ans 105
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Ho chuka Hai means what?
‘It has been done ‘
@factorialacademy dhanyavaad
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.
Homework ans is 15C2=105
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Hw ans 105?
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Is the answer 105?
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Bhai ek video skew lines par bana do pls
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
Hw question ans 105
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105
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@factorialacademy can you make a video on important models of calculus?