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Question Number 173992 by dragan91 last updated on 22/Jul/22

Commented by mr W last updated on 22/Jul/22

no answer seems to be correct. i got 5456.

$${no}\:{answer}\:{seems}\:{to}\:{be}\:{correct}.\:{i}\:{got} \\ $$$$\mathrm{5456}. \\ $$

Answered by mr W last updated on 22/Jul/22

x_1 +x_2 +x_3 =n has C_2 ^(n+2) solutions with x_i ≥0. total number of solutions for 0≤n≤30: Σ_(n=0) ^(30) C_2 ^(n+2) =Σ_(n=0) ^(30) (((n+2)(n+1))/2) =Σ_(n=0) ^(30) (((n^2 +3n+2))/2) =(1/2)(((30×31×61)/6)+3×((30×31)/2)+31×2) =5456

$${x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} ={n}\:{has}\:{C}_{\mathrm{2}} ^{{n}+\mathrm{2}} \:{solutions} \\ $$$${with}\:{x}_{{i}} \geqslant\mathrm{0}. \\ $$$${total}\:{number}\:{of}\:{solutions}\:{for}\:\mathrm{0}\leqslant{n}\leqslant\mathrm{30}: \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\mathrm{30}} {\sum}}{C}_{\mathrm{2}} ^{{n}+\mathrm{2}} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\mathrm{30}} {\sum}}\frac{\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\mathrm{30}} {\sum}}\frac{\left({n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}\right)}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{30}×\mathrm{31}×\mathrm{61}}{\mathrm{6}}+\mathrm{3}×\frac{\mathrm{30}×\mathrm{31}}{\mathrm{2}}+\mathrm{31}×\mathrm{2}\right) \\ $$$$=\mathrm{5456} \\ $$

Commented by dragan91 last updated on 22/Jul/22

n starts from 3 since x_i ∈N

$$\mathrm{n}\:\mathrm{starts}\:\mathrm{from}\:\mathrm{3}\:\mathrm{since}\:\mathrm{x}_{{i}} \in\mathrm{N} \\ $$

Commented by dragan91 last updated on 22/Jul/22

Number of positive integers if equation x_1 +x_2 +x_3 =n for n≥3 is C_2 ^(n−1)

$$\mathrm{Number}\:\mathrm{of}\:\mathrm{positive}\:\mathrm{integers}\:\mathrm{if}\:\mathrm{equation} \\ $$$$\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} +\mathrm{x}_{\mathrm{3}} =\mathrm{n}\:\mathrm{for}\:\mathrm{n}\geqslant\mathrm{3}\:\mathrm{is}\:\mathrm{C}_{\mathrm{2}} ^{\mathrm{n}−\mathrm{1}} \\ $$$$ \\ $$

Commented by dragan91 last updated on 22/Jul/22

At first I get 4060

$$\mathrm{At}\:\mathrm{first}\:\mathrm{I}\:\mathrm{get}\:\mathrm{4060} \\ $$

Commented by mr W last updated on 22/Jul/22

to me 0∈N. besides the question states 0≤x_1 +x_2 +x_3 ≤30.

$${to}\:{me}\:\mathrm{0}\in\mathbb{N}.\:{besides}\:{the}\:{question}\:{states} \\ $$$$\mathrm{0}\leqslant{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} \leqslant\mathrm{30}. \\ $$

Commented by mr W last updated on 22/Jul/22

Commented by dragan91 last updated on 22/Jul/22

Ok.For me 0 is not natural. I will check with other people. I think the given answers is all wrong. Thanks.

$$\mathrm{Ok}.\mathrm{For}\:\mathrm{me}\:\mathrm{0}\:\mathrm{is}\:\mathrm{not}\:\mathrm{natural}.\:\mathrm{I}\:\mathrm{will}\:\mathrm{check}\:\mathrm{with} \\ $$$$\mathrm{other}\:\mathrm{people}.\:\mathrm{I}\:\mathrm{think}\:\mathrm{the}\:\mathrm{given}\:\mathrm{answers}\:\mathrm{is}\:\mathrm{all}\:\mathrm{wrong}.\:\mathrm{Thanks}. \\ $$$$ \\ $$

Commented by Tawa11 last updated on 22/Jul/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Commented by mr W last updated on 22/Jul/22

it is just a question of definition if 0 is a natural number or not. in Germany it′s even differently defined in different federal states.

$${it}\:{is}\:{just}\:{a}\:{question}\:{of}\:{definition} \\ $$$${if}\:\mathrm{0}\:{is}\:{a}\:{natural}\:{number}\:{or}\:{not}. \\ $$$${in}\:{Germany}\:{it}'{s}\:{even}\:{differently} \\ $$$${defined}\:{in}\:{different}\:{federal}\:{states}. \\ $$

Commented by dragan91 last updated on 22/Jul/22

thanks

$$\mathrm{thanks} \\ $$

Commented by MJS_new last updated on 23/Jul/22

once we used N={1, 2, 3, ...} and N_0 ={0, 1, 2, 3, ...} now most people use N={0, 1, 2, 3, ...} and N^★ ={1, 2, 3, ...}

$$\mathrm{once}\:\mathrm{we}\:\mathrm{used} \\ $$$$\mathbb{N}=\left\{\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:...\right\}\:\mathrm{and}\:\mathbb{N}_{\mathrm{0}} =\left\{\mathrm{0},\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:...\right\} \\ $$$$\mathrm{now}\:\mathrm{most}\:\mathrm{people}\:\mathrm{use} \\ $$$$\mathbb{N}=\left\{\mathrm{0},\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:...\right\}\:\mathrm{and}\:\mathbb{N}^{\bigstar} =\left\{\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:...\right\} \\ $$

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