combinatorics: what is the coefficient of (a^2)(b^3)(c) in (2a - b + 3c)^6?
Accepted Solution
A:
Answer:Hence, the coefficient of a²b³c = -720.Step-by-step explanation:As from the question,The general formula to find the coefficient is given by Binomial theorem:That is, the coefficient of [tex]x^{\alpha}\cdot y^{\beta}\cdot z^{\gamma}[/tex] in (x + y + z)ⁿ is given by:[tex]\frac{n!}{\alpha ! \cdot \beta ! \cdot \gamma !} (x)^{\alpha} \cdot (y)^{\beta} \cdot (z)^{\gamma}[/tex]Now,From the question we have[tex](2a-b+3c)^{6}[/tex] having n = 6∴ x = 2ay = -bz = 3cNow,The coefficient of a²b³c, that isα = 2β = 3γ = 1Therefore the coefficient of a²b³c =[tex]= \frac{6!}{2 ! \cdot 3 ! \cdot 1 !} (2a)^{2} \cdot (-b)^{3} \cdot (3c)^{1}[/tex][tex]= \frac{6!}{2 ! \cdot 3 ! \cdot 1 !} 4(a)^{2} \cdot (-b)^{3} \cdot (3c)[/tex]= -720 a²b³cHence, the coefficient of a²b³c = -720.