how-many-terms-contain-ab-2-c-3-in-a-2b-3c-4d-2-5e-3-10-

Question Number 155109 by qaz last updated on 25/Sep/21

how many terms contain “ {ab}^{2} c^{3}^{″} in {(a + 2 b + 3 c + {4 d}^{2} + {5 e}^{3})}^{10} ?

Answered by mr W last updated on 26/Sep/21

{(a + 2 b + 3 c + 4 d^{2} + 5 e^{3})}^{10}

= \underset{0 ⩽ α, β, γ, δ, ϵ ⩽ 10}{\sum^{α + β + γ + δ + ϵ = 10}} (\begin{matrix} 10 \\ α, β, γ, δ, ϵ \end{matrix}) a^{α} {(2 b)}^{β} {(3 c)}^{γ} {(4 d^{2})}^{δ} {(5 e^{3})}^{ϵ}

t e r m s c o n t a i n i n g {a b}^{2} c^{3} f u l f i l l

α + β + γ + δ + ϵ = 10 \dots (i)

w i t h

α ⩾ 1 \Rightarrow x + x^{2} + x^{3} + \dots = x (1 + x + x^{2} + \dots)

β ⩾ 2 \Rightarrow x^{2} + x^{3} + x^{4} \dots = x^{2} (1 + x + x^{2} + \dots)

γ ⩾ 3 \Rightarrow x^{3} + x^{4} + x^{5} + \dots = x^{3} (1 + x + x^{2} + \dots)

δ ⩾ 0 \Rightarrow 1 + x + x^{2} + x^{3} + \dots = (1 + x + x^{2} + \dots)

ϵ ⩾ 0 \Rightarrow 1 + x + x^{2} + x^{3} + \dots = (1 + x + x^{2} + \dots)

x^{6} {(1 + x + x^{2} + \dots)}^{5} = \frac{x^{6}}{{(1 - x)}^{5}} = \underset{k = 0}{\sum^{\infty}} C_{4}^{k + 4} x^{k + 6}

c o e f . o f x^{10} i s C_{4}^{4 + 4} = C_{4}^{8} = 70

t h a t m e a n s (i) h a s 70 s o l u t i o n s

s a t i s f y i n g g i v e n c o n d i t i o n s . i . e .

t h e r e a r e 70 t e r m s c o n t a i n i n g {a b}^{2} c^{3} .

Commented by Tawa11 last updated on 26/Sep/21

great sir

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