pnc

Why it is coeffiecient of that term (I am just not getting it)
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24 Replies
iTeachChem Helper
@Apu
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BlindSniper (BS)
well it kind of is a set rule just like pascal's triangle they just found a way to expand (a+b)^n and this is the method they created and it worked so it's always been like that
SirLancelotDuLac
Write (x+y)ⁿ as (x+y)(x+y)...(x+y) then you will figure it out.
SirLancelotDuLac
Hmm that works too.
BlindSniper (BS)
but in case of pascal's triangle the coefficients are always 1 for x and y in binomial theorem there's questions for coefficients with x and y having coefficients which is another story for now but this is the explanation, I guess
BlindSniper (BS)
oh a blackboard epic shi
Gaurav
GauravOP4mo ago
your blackboard is so shiny my teacher explained like "agar total terms n hai and usko do bando mein bantna hai r and n-r ki amounts me then we get the factorial from pnc and from there we get the coefficient formula of nCr" @SirLancelotDuLac
BlindSniper (BS)
r and r-1?
Gaurav
GauravOP4mo ago
corrected ...
SirLancelotDuLac
Yeah. Look at it this way (x+y)(x+y)...(x+y) mein n brackets hain. Now agar isse expand karke likhein toh $x^{a}y^{n-a}$ tabh aayega jabh n brackets mein se a brackets se x aur baakiyon se y multiply karein. So the number of ways for that is nCr
TeXit
TeXit4mo ago
SirLancelotDuLac
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SirLancelotDuLac
The pascal triangle method is another way. Define $a{n,r}$ as co-effecient of $x^{r}$ in $(x+y)^{n}$. Not necessary but if you are interested try to solve this recursion: $a{n,r}=a{n-1,r-1}+a{n-1,r}$
TeXit
TeXit4mo ago
SirLancelotDuLac
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Gaurav
GauravOP4mo ago
thoda thoda samaj aya but ithink yaad hi kr leta hu :Saul_Goodman:
BlindSniper (BS)
well I mean that much sort of obvious from the expression itself he just told you in words
iTeachChem
iTeachChem4mo ago
yaad karne se life mushkil hogi abi sorted?
Gaurav
GauravOP4mo ago
Nhi wait ek Baar phirse pdhata sable explanations
iTeachChem
iTeachChem4mo ago
Cool
Gaurav
GauravOP4mo ago
+solved @BlindSniper (BS) @SirLancelotDuLac
iTeachChem Helper
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