Jtdylktuy

0 $begingroup$ Number of positive unequal integer solution of $x_{1}+x_{2}+x_{3}+x_{4}=20,$ is Try: Total number of positive integer solution is $displaystyle binom{20-1}{4-1}=binom{19}{3}.$ (Using star and bar method) Now for unequal integer solution, First we will find Total positive integer solution. And subtract the cases where any three are equal and any two are equal and all are equal. So Total number of positive integer solution is $displaystyle binom{20-1}{4-1}=binom{19}{3}.$ (Using star and bar method) And number of positive integer solution in which all are equal i. e $x_{1}=x_{2}=x_{3}=x_{4}.$ Which is only one quadraplets. Now if all any three are equal i. e $x_{1}=x_{2}=x_{3}$ . Then $3x_{1}+x_{4}=20.$ we have $6times 4=24.$ If any two are equal. i. e $x_{1}=x_{2}.$