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Question Number 103903 by mr W last updated on 18/Jul/20

a 100 cm long rod should be divided  into 3 parts. the length of each part  in cm should be integer. in how   many different ways can this be  done?

$${a}\:\mathrm{100}\:{cm}\:{long}\:{rod}\:{should}\:{be}\:{divided} \\ $$$${into}\:\mathrm{3}\:{parts}.\:{the}\:{length}\:{of}\:{each}\:{part} \\ $$$${in}\:{cm}\:{should}\:{be}\:{integer}.\:{in}\:{how}\: \\ $$$${many}\:{different}\:{ways}\:{can}\:{this}\:{be} \\ $$$${done}? \\ $$

Commented by bobhans last updated on 18/Jul/20

in parts alowed the same length?

$${in}\:{parts}\:{alowed}\:{the}\:{same}\:{length}? \\ $$

Commented by mr W last updated on 18/Jul/20

the only condition is: the length  must be integer. so the parts may  also have equal length.

$${the}\:{only}\:{condition}\:{is}:\:{the}\:{length} \\ $$$${must}\:{be}\:{integer}.\:{so}\:{the}\:{parts}\:{may} \\ $$$${also}\:{have}\:{equal}\:{length}. \\ $$

Commented by bobhans last updated on 18/Jul/20

it same x_1 +x_2 +x_3  = 100 , with x_i ∈Z^+

$${it}\:{same}\:{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} \:=\:\mathrm{100}\:,\:{with}\:{x}_{{i}} \in\mathbb{Z}^{+} \\ $$

Commented by mr W last updated on 18/Jul/20

not exactly the same.   e.g. 2+5+93 or 5+2+93 or 93+2+5 are  different solutions for your  equation, but they represent the one  and the same way to divide the rod.  the parts are not labeled.

$${not}\:{exactly}\:{the}\:{same}.\: \\ $$$${e}.{g}.\:\mathrm{2}+\mathrm{5}+\mathrm{93}\:{or}\:\mathrm{5}+\mathrm{2}+\mathrm{93}\:{or}\:\mathrm{93}+\mathrm{2}+\mathrm{5}\:{are} \\ $$$${different}\:{solutions}\:{for}\:{your} \\ $$$${equation},\:{but}\:{they}\:{represent}\:{the}\:{one} \\ $$$${and}\:{the}\:{same}\:{way}\:{to}\:{divide}\:{the}\:{rod}. \\ $$$${the}\:{parts}\:{are}\:{not}\:{labeled}. \\ $$

Commented by bobhans last updated on 18/Jul/20

o oo yes agree

$${o}\:{oo}\:{yes}\:{agree}\: \\ $$

Answered by mr W last updated on 18/Jul/20

METHOD I  say the lengthes of the parts are  a,b,c with a≤b≤c.  a+b+c=100  100=a+b+c≥3a  ⇒a≤((100)/3) ⇒a≤33  with a=1:  99=b+c≥2b  ⇒b≤((99)/2) ⇒b≤49  ⇒b=1,2,...,49 ⇒49 solutions  with a=2:  98=b+c≥2b  ⇒b≤49  ⇒b=2,3,...,49 ⇒48 solutions  with a=3:  97=b+c≥2b  ⇒b≤((97)/2) ⇒b≤48  ⇒b=3,4,...,48 ⇒46 solutions  with a=4:  96=b+c≥2b  ⇒b≤48  ⇒b=4,5,...,48 ⇒45 solutions  .....  with a=33:  67=b+c≥2b  ⇒b≤((67)/2) ⇒b≤33  ⇒b=33 ⇒1 solution    total number of solutions:  1+4+7+...+46+49  +3+6+9+...+45+48  =((17×(1+49))/2)+((16×(3+48))/2)  =833

$$\boldsymbol{{METHOD}}\:\boldsymbol{{I}} \\ $$$${say}\:{the}\:{lengthes}\:{of}\:{the}\:{parts}\:{are} \\ $$$${a},{b},{c}\:{with}\:{a}\leqslant{b}\leqslant{c}. \\ $$$${a}+{b}+{c}=\mathrm{100} \\ $$$$\mathrm{100}={a}+{b}+{c}\geqslant\mathrm{3}{a} \\ $$$$\Rightarrow{a}\leqslant\frac{\mathrm{100}}{\mathrm{3}}\:\Rightarrow{a}\leqslant\mathrm{33} \\ $$$${with}\:{a}=\mathrm{1}: \\ $$$$\mathrm{99}={b}+{c}\geqslant\mathrm{2}{b} \\ $$$$\Rightarrow{b}\leqslant\frac{\mathrm{99}}{\mathrm{2}}\:\Rightarrow{b}\leqslant\mathrm{49} \\ $$$$\Rightarrow{b}=\mathrm{1},\mathrm{2},...,\mathrm{49}\:\Rightarrow\mathrm{49}\:{solutions} \\ $$$${with}\:{a}=\mathrm{2}: \\ $$$$\mathrm{98}={b}+{c}\geqslant\mathrm{2}{b} \\ $$$$\Rightarrow{b}\leqslant\mathrm{49} \\ $$$$\Rightarrow{b}=\mathrm{2},\mathrm{3},...,\mathrm{49}\:\Rightarrow\mathrm{48}\:{solutions} \\ $$$${with}\:{a}=\mathrm{3}: \\ $$$$\mathrm{97}={b}+{c}\geqslant\mathrm{2}{b} \\ $$$$\Rightarrow{b}\leqslant\frac{\mathrm{97}}{\mathrm{2}}\:\Rightarrow{b}\leqslant\mathrm{48} \\ $$$$\Rightarrow{b}=\mathrm{3},\mathrm{4},...,\mathrm{48}\:\Rightarrow\mathrm{46}\:{solutions} \\ $$$${with}\:{a}=\mathrm{4}: \\ $$$$\mathrm{96}={b}+{c}\geqslant\mathrm{2}{b} \\ $$$$\Rightarrow{b}\leqslant\mathrm{48} \\ $$$$\Rightarrow{b}=\mathrm{4},\mathrm{5},...,\mathrm{48}\:\Rightarrow\mathrm{45}\:{solutions} \\ $$$$..... \\ $$$${with}\:{a}=\mathrm{33}: \\ $$$$\mathrm{67}={b}+{c}\geqslant\mathrm{2}{b} \\ $$$$\Rightarrow{b}\leqslant\frac{\mathrm{67}}{\mathrm{2}}\:\Rightarrow{b}\leqslant\mathrm{33} \\ $$$$\Rightarrow{b}=\mathrm{33}\:\Rightarrow\mathrm{1}\:{solution} \\ $$$$ \\ $$$${total}\:{number}\:{of}\:{solutions}: \\ $$$$\mathrm{1}+\mathrm{4}+\mathrm{7}+...+\mathrm{46}+\mathrm{49} \\ $$$$+\mathrm{3}+\mathrm{6}+\mathrm{9}+...+\mathrm{45}+\mathrm{48} \\ $$$$=\frac{\mathrm{17}×\left(\mathrm{1}+\mathrm{49}\right)}{\mathrm{2}}+\frac{\mathrm{16}×\left(\mathrm{3}+\mathrm{48}\right)}{\mathrm{2}} \\ $$$$=\mathrm{833} \\ $$

Commented by bobhans last updated on 18/Jul/20

cooll....nice

$${cooll}....{nice} \\ $$

Commented by mr W last updated on 18/Jul/20

METHOD II  say the lengthes of the parts are  a,b,c with a≤b≤c.  a+b+c=100   ...(i)  number of solutions is the coeff. of  x^(100)  term of following generating  function  (x^3 /((1−x)(1−x^2 )(1−x^3 )))  which is 833.

$$\boldsymbol{{METHOD}}\:\boldsymbol{{II}} \\ $$$${say}\:{the}\:{lengthes}\:{of}\:{the}\:{parts}\:{are} \\ $$$${a},{b},{c}\:{with}\:{a}\leqslant{b}\leqslant{c}. \\ $$$$\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}=\mathrm{100}\:\:\:...\left({i}\right) \\ $$$${number}\:{of}\:{solutions}\:{is}\:{the}\:{coeff}.\:{of} \\ $$$${x}^{\mathrm{100}} \:{term}\:{of}\:{following}\:{generating} \\ $$$${function} \\ $$$$\frac{{x}^{\mathrm{3}} }{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\mathrm{1}−{x}^{\mathrm{3}} \right)} \\ $$$${which}\:{is}\:\mathrm{833}. \\ $$

Commented by mr W last updated on 18/Jul/20

no sir!  (a+b+c)^(100)  has no meaning.  it is the coefficient of x^(100)  in the  generating function  (x^3 /((1−x)(1−x^2 )(1−x^3 )))

$${no}\:{sir}! \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{100}} \:{has}\:{no}\:{meaning}. \\ $$$${it}\:{is}\:{the}\:{coefficient}\:{of}\:{x}^{\mathrm{100}} \:{in}\:{the} \\ $$$${generating}\:{function} \\ $$$$\frac{{x}^{\mathrm{3}} }{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\mathrm{1}−{x}^{\mathrm{3}} \right)} \\ $$

Commented by mr W last updated on 18/Jul/20

Commented by bemath last updated on 18/Jul/20

coeff of x^(100)  from (a+b+c)^(100)   sir?

$${coeff}\:{of}\:{x}^{\mathrm{100}} \:{from}\:\left({a}+{b}+{c}\right)^{\mathrm{100}} \\ $$$${sir}? \\ $$

Commented by mr W last updated on 18/Jul/20

it means in how many ways can the  number 100 be divided into 3 parts.

$${it}\:{means}\:{in}\:{how}\:{many}\:{ways}\:{can}\:{the} \\ $$$${number}\:\mathrm{100}\:{be}\:{divided}\:{into}\:\mathrm{3}\:{parts}. \\ $$

Commented by bemath last updated on 18/Jul/20

if number 200 be divided  into 3 parts then equal to  coeff of x^(200)  sir?

$${if}\:{number}\:\mathrm{200}\:{be}\:{divided} \\ $$$${into}\:\mathrm{3}\:{parts}\:{then}\:{equal}\:{to} \\ $$$${coeff}\:{of}\:{x}^{\mathrm{200}} \:{sir}? \\ $$

Commented by mr W last updated on 18/Jul/20

yes

$${yes} \\ $$

Commented by bemath last updated on 18/Jul/20

thank you sir. i will learn

$${thank}\:{you}\:{sir}.\:{i}\:{will}\:{learn}\: \\ $$

Commented by prakash jain last updated on 18/Jul/20

Some more ideas around this topic in Wikipedia https://en.m.wikipedia.org/wiki/Partition_(number_theory)

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