SHAIK AASIF AHAMED
Last Activity: 10 Years ago
Hello student,
Please find the answer to your question below
Let A,B,C donatex1,x2,x3coins withxi≥0.
Then A/Q∑xi=10.....(1)
At first lets find all the solutions of this equation in integers.
The no. of such solution is12C2
Now we will find the no. of solution in whichx1≥7,(these solutions cant be considered),
To find this lets replacex1byx+6wherex≥1putting this into 1 we havex+x2+x3=4we will find no. of such solns.5C2
In this way we will find the other cases which are not possible.
Namely whenx2≥8in this case we have4C2solutions , and the last one whenx3≥9then we have3C2 solutions.
All the cases which cant be posiible are disjoint implying
the total no. of solution =(total no. of cases)-(cases not possible).
=12C2−5C2−4C2−3C2
=47