The number of integers x, y, z, w such that x+y+z+w=20 and x, y, z, w ≥-1, is

Question

The number of integers x, y, z, w such that x+y+z+w=20 and x, y, z, w ≥-1, is (A)  24 C3 (B)  25 C3 (C)  26 C3 (D)  27 C3

Tardigrade Admin · Accepted Answer

Put x=a-1, y=b-1, z=c-1, w=d-1, then a, b, c, d ≥ 0 and (a-1)+(b-1)+(c-1)+(d-1)=20 ⇒ a+b+c+d=24 The number of non-negative integral solutions of this equation is 24+4-1 C4-1= 27 C3

Q. The number of integers $x, y, z, w$ such that $x + y + z + w = 20$ and $x, y, z, w \geq - 1$ , is

$^{24} C_{3}$

$^{25} C_{3}$

$^{26} C_{3}$

$^{27} C_{3}$

Put $x = a - 1, y = b - 1, z = c - 1, w = d - 1$ , then $a, b, c, d \geq 0$ and $(a - 1) + (b - 1) + (c - 1) + (d - 1) = 20$ $\Rightarrow a + b + c + d = 24$
The number of non-negative integral solutions of this equation is
$^{24 + 4 - 1} C_{4 - 1} =^{27} C_{3}$

Q. The number of integers x,y,z,w such that x+y+z+w=20 and x,y,z,w≥−1, is

24C3​

25C3​

26C3​

27C3​