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Question Number 206156 by cortano21 last updated on 08/Apr/24

Commented by mr W last updated on 08/Apr/24

$w h a t i t h i n k i s c l e a r :$ $i n t h e p a c k a g e t h e r e m u s t b e e x a c t l y$ $o n e e g g . t h e r e m u s t b e a t l e a s t t h r e e$ $l e t t u c e s . t h e r e m u s t b e a t l e a s t o n e$ $t o f u, a t l e a s t o n e p o t a t o . b i t t e r m e l o n$ $m a y b e o m i t t e d . c a b b a g e m a y b e$ $o m i t t e d .$ $w h a t i s u n c l e a r :$ $t h e m a x i m u m n u m b e r o f b i t t e r m e l o n$ $a n d c a b b a g e i s 1 . d o e s i t m e a n :$ $t o t a l l y t h e r e i s a t m o s t o n e b i t t e r$ $m e l o n o r o n e c a b b a g e ?$ $i . e . b i t t e r m e l o n + c a b b a g e ⩽ 1 .$ $o r$ $t h e r e i s a t m o s t o n e b i t t e r m e l o n$ $a n d a t m o s t o n e c a b b a g e ?$ $i . e . b i t t e r m e l o n ⩽ 1 a n d c a b b a g e ⩽ 1 .$
Commented by cortano21 last updated on 09/Apr/24

$y e s s i r$
Commented by mr W last updated on 09/Apr/24

$i a s k e d w h a t i s m e a n t, o p t i o n 1 o r$ $o p t i o n 2 . y o u a n s w e r e d y e s . t h i s i s$ $n o t a p r o p e r a n s w e r .$
Commented by mr W last updated on 09/Apr/24

${l e t}^{'} s a s s u m e t h a t o p t i o n 1 i s m e a n t,$ $i . e . b i t t e r m e l o n + c a b b a g e ⩽ 1 .$
	Answered by mr W last updated on 09/Apr/24

	$L = n u m b e r o f l e t t u c e s, L ⩾ 3$ $T = n u m b e r o f t o f u, T ⩾ 1$ $E = n u m b e r o f e g g s, E = 1$ $P = n u m b e r o f p o t a t o e s, P ⩾ 1$ $C = n u m b e r o f c a b b a g e, 0 ⩽ C$ $B = n u m b e r o f b i t t e r l e m o n s, 0 ⩽ B$ $b u t K = B + C ⩽ 1$ $L + T + E + P + C + B = 10$ $L + T + P + K = 9$ $\underset{L}{(x^{3} + x^{4} + . . .)} \underset{T, P}{{(x + x^{2} + x^{3} + . . .)}^{2}} \underset{K = B + C}{(1 + 2 x)}$ $= (1 + 2 x) x^{5} {(1 + x + x^{2} + x^{3} + . . .)}^{3}$ $= \frac{(1 + 2 x) x^{5}}{{(1 - x)}^{3}} = (1 + 2 x) x^{5} \underset{k = 0}{\sum^{\infty}} C_{2}^{k + 2} x^{k}$ $f o r x^{9} t e r m :$ $k = 4 : C_{2}^{6} = 15$ $k = 3 : 2 C_{2}^{5} = 20$ $\Rightarrow c o e f . o f x^{9} i s 15 + 20 = 35$ $t h a t m e a n s t h e r e a r e 35 w a y s t o$ $c h o o s e t h e c o n t e n t o f f o o d p a c k a g e .$
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