How many integer solutions are there to the equation x1 + x2 + x3 + x4 = 132 pro

Question

How many integer solutions are there to the equation x1 + x2 + x3 + x4 = 132 provided that x1 > 0, and x2, x3, x4 ge 0? What if we add the restriction that x4

Explanation / Answer

Case 1: x2=0 , x3,x4>0. therefore x1+x3+x4=132. no of solution possible - > 131C2 = 131!/(2!)(129!) = 131*130/2 = 8515 same case is possible when x3=0 , x1,x4>0 and x4=0 , x3,x2>0.. thus number of solutions ..... 3*8515 = 25545 CASE 2 : when X2,x3 = 0 ...x4>0 x1+x4=132. no of solutions -> 131C1 = 131 Again same case is possible when X4,x3 = 0 ...x2>0 and X2,x4= 0 ...x2>0 no of solutions -> 3*131 = 393. Case 3 : X2,x3,x4 = 0 x1+x2+x3+x4=132: no of soltions -> 131C3 = 131*130*129/6 = 366145 therefore total number of solutions -> 25545 + 393 +366145 = 392083 (Ans)

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.