Question

Number of integral solutions to the equation x + y + z = 21, where x ≥ 1, y ≥ 3, z ≥ 4, is equal to ______.

Comments

14C2  = 14! /2! (14-2)!
Which is given by formula n+1C r-1
Where n = x+y+z= 8 and r = no. Of given variables in equation like x, y, z

---

Answer is correct, formula for finding no. of non-negative integral solution is ^{n+r-1}Cr-1 (not ^{n+1}Cr-1).

Answer 1

\(x + y +z = 21, x \ge1, y \ge 3, z\ge 4\)

⇒ \((x - 1) + (y - 3) + (z-4) = 21 - 8 = 13\)

⇒ \(\alpha + \beta + \gamma = 13, \alpha \ge 0, \beta \ge 0, \gamma \ge 0\)

where \(\alpha = x-1, \beta = y - 3\text{ & } \gamma = z - 4\)

\(\therefore \) No. of integral solution = \(^{n+r-1}C_{r-1}\)

where n = 13, r = number of variables = 3

 \(\therefore \) No. of integral solution = \(^{13 +3 -1}C_{3 -1} =\, ^{15}C_2\)

\(= \frac{15 \times 14}{2} \)

\(= 105\)

Answer 2

\(^{15}C_2 = \frac{15 \times 14}2 = 105\)

Comments

Your answer is wrong it should be 14C2 = 91

---

Answer is correct, formula for finding no. of non-negative integral solution is ^{n+r-1}Cr-1 (not ^{n+1}Cr-1).