If x, y, z, w ∈ N, such that each of them gives remainder 3 when divided by 5 , then number of solutions of the equation x+y+z+w=62, is equal to

Question

If x, y, z, w ∈ N, such that each of them gives remainder 3 when divided by 5 , then number of solutions of the equation x+y+z+w=62, is equal to (A) 13 C 3 (B)  9 C 3 (C)  12 C 3 (D)  12 C 2

Tardigrade Admin · Accepted Answer

x = 5a + 3 y = 5b + 3 z = 5c + 3 w = 5d + 3 x + y + z + w =5( a + b + c + d )+12=62= a + b + c + d =10 text so 10+4- 4 C 4-1= 13 C 3

Q. If $x, y, z, w \in N$ , such that each of them gives remainder 3 when divided by 5 , then number of solutions of the equation $x + y + z + w = 62$ , is equal to

$^{13} C_{3}$

$^{9} C_{3}$

$^{12} C_{3}$

$^{12} C_{2}$

$x = 5 a + 3$
$y = 5 b + 3$
$z = 5 c + 3$
$w = 5 d + 3$
$x + y + z + w = 5 (a + b + c + d) + 12 = 62 = {a + b + c + d = 10}$
$so^{10 + 4 -^{4} C_{4 - 1}} =^{13} C_{3}$

Q. If x,y,z,w∈N, such that each of them gives remainder 3 when divided by 5 , then number of solutions of the equation x+y+z+w=62, is equal to

13C3​

9C3​

12C3​

12C2​