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Question-206156

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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
what i think is clear: in the package there must be exactly one egg. there must be at least three lettuces. there must be at least one tofu, at least one potato. bitter melon may be omitted. cabbage may be omitted. what is unclear: the maximum number of bitter melon and cabbage is 1. does it mean: totally there is at most one bitter melon or one cabbage? i.e. bitter melon+cabbage≤1. or there is at most one bitter melon and at most one cabbage? i.e. bitter melon≤1 and cabbage≤1.
$${what}\:{i}\:{think}\:{is}\:{clear}: \\ $$$${in}\:{the}\:{package}\:{there}\:{must}\:{be}\:{exactly} \\ $$$${one}\:{egg}.\:{there}\:{must}\:{be}\:{at}\:{least}\:{three} \\ $$$${lettuces}.\:{there}\:{must}\:{be}\:{at}\:{least}\:{one} \\ $$$${tofu},\:{at}\:{least}\:{one}\:{potato}.\:{bitter}\:{melon} \\ $$$${may}\:{be}\:{omitted}.\:{cabbage}\:{may}\:{be} \\ $$$${omitted}. \\ $$$${what}\:{is}\:{unclear}: \\ $$$${the}\:{maximum}\:{number}\:{of}\:{bitter}\:{melon} \\ $$$${and}\:{cabbage}\:{is}\:\mathrm{1}.\:{does}\:{it}\:{mean}: \\ $$$${totally}\:{there}\:{is}\:{at}\:{most}\:{one}\:{bitter} \\ $$$${melon}\:{or}\:{one}\:{cabbage}? \\ $$$${i}.{e}.\:{bitter}\:{melon}+{cabbage}\leqslant\mathrm{1}. \\ $$$${or} \\ $$$${there}\:{is}\:{at}\:{most}\:{one}\:{bitter}\:{melon} \\ $$$${and}\:{at}\:{most}\:{one}\:{cabbage}? \\ $$$${i}.{e}.\:{bitter}\:{melon}\leqslant\mathrm{1}\:{and}\:{cabbage}\leqslant\mathrm{1}. \\ $$
Commented by cortano21 last updated on 09/Apr/24
yes sir
$${yes}\:{sir} \\ $$
Commented by mr W last updated on 09/Apr/24
i asked what is meant, option 1 or option 2. you answered yes. this is not a proper answer.
$${i}\:{asked}\:{what}\:{is}\:{meant},\:{option}\:\mathrm{1}\:{or} \\ $$$${option}\:\mathrm{2}.\:{you}\:{answered}\:{yes}.\:{this}\:{is} \\ $$$${not}\:{a}\:{proper}\:{answer}. \\ $$
Commented by mr W last updated on 09/Apr/24
let′s assume that option 1 is meant, i.e. bitter melon+cabbage≤1.
$${let}'{s}\:{assume}\:{that}\:{option}\:\mathrm{1}\:{is}\:{meant}, \\ $$$${i}.{e}.\:{bitter}\:{melon}+{cabbage}\leqslant\mathrm{1}. \\ $$
Answered by mr W last updated on 09/Apr/24
L=number of lettuces, L≥3 T=number of tofu, T≥1 E=number of eggs, E=1 P=number of potatoes, P≥1 C=number of cabbage, 0≤C B=number of bitter lemons, 0≤B but K=B+C≤1 L+T+E+P+C+B=10 L+T+P+K=9 (x^3 +x^4 +...)_(L) (x+x^2 +x^3 +...)^2 _(T, P) (1+2x)_(K=B+C) =(1+2x)x^5 (1+x+x^2 +x^3 +...)^3 =(((1+2x)x^5 )/((1−x)^3 ))=(1+2x)x^5 Σ_(k=0) ^∞ C_2 ^(k+2) x^k for x^9 term: k=4: C_2 ^6 =15 k=3: 2C_2 ^5 =20 ⇒ coef. of x^9 is 15+20=35 that means there are 35 ways to choose the content of food package.
$${L}={number}\:{of}\:{lettuces},\:{L}\geqslant\mathrm{3} \\ $$$${T}={number}\:{of}\:{tofu},\:{T}\geqslant\mathrm{1} \\ $$$${E}={number}\:{of}\:{eggs},\:{E}=\mathrm{1} \\ $$$${P}={number}\:{of}\:{potatoes},\:{P}\geqslant\mathrm{1} \\ $$$${C}={number}\:{of}\:{cabbage},\:\mathrm{0}\leqslant{C} \\ $$$${B}={number}\:{of}\:{bitter}\:{lemons},\:\mathrm{0}\leqslant{B} \\ $$$$\:\:\:\:\:\:\:\:{but}\:{K}={B}+{C}\leqslant\mathrm{1} \\ $$$${L}+{T}+{E}+{P}+{C}+{B}=\mathrm{10} \\ $$$${L}+{T}+{P}+{K}=\mathrm{9} \\ $$$$\underset{{L}} {\left({x}^{\mathrm{3}} +{x}^{\mathrm{4}} +…\right)}\underset{{T},\:{P}} {\left({x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…\right)^{\mathrm{2}} }\underset{{K}={B}+{C}} {\left(\mathrm{1}+\mathrm{2}{x}\right)} \\ $$$$=\left(\mathrm{1}+\mathrm{2}{x}\right){x}^{\mathrm{5}} \left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…\right)^{\mathrm{3}} \\ $$$$=\frac{\left(\mathrm{1}+\mathrm{2}{x}\right){x}^{\mathrm{5}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }=\left(\mathrm{1}+\mathrm{2}{x}\right){x}^{\mathrm{5}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{2}} ^{{k}+\mathrm{2}} {x}^{{k}} \\ $$$${for}\:{x}^{\mathrm{9}} \:{term}: \\ $$$${k}=\mathrm{4}:\:{C}_{\mathrm{2}} ^{\mathrm{6}} =\mathrm{15} \\ $$$${k}=\mathrm{3}:\:\mathrm{2}{C}_{\mathrm{2}} ^{\mathrm{5}} =\mathrm{20} \\ $$$$\Rightarrow\:{coef}.\:{of}\:{x}^{\mathrm{9}} \:{is}\:\mathrm{15}+\mathrm{20}=\mathrm{35} \\ $$$${that}\:{means}\:{there}\:{are}\:\mathrm{35}\:{ways}\:{to} \\ $$$${choose}\:{the}\:{content}\:{of}\:{food}\:{package}. \\ $$

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