If x, y, and z are whole numbers such that x is greater than or equal to y, then how many solutions are possible for the equation x + y + z = 36?

361

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

323

No worries! We‘ve got your back. Try BYJU‘S free classes today!

382

No worries! We‘ve got your back. Try BYJU‘S free classes today!

342

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 361
The total number of solutions for x + y + z = 36, if x, y and z are whole numbers is given by
$^{36 + 3 - 1} C_{3 - 1} =^{38} C_{2} = 703$ .
The number of solutions where x = y will be 19
(from (x, y) = (0, 0) to (18, 18)).
The number of solutions where x is not equal to y
= 703 - 19 = 684
Among these 684 solutions, half will have x > y and,
the rest will have y > x.
Hence, the total number of solutions where $x \geq y = 19 + \frac{684}{2} = 361$

Suggest Corrections

Similar questions

Q. How many natural number solutions are possible for equation below ?
x + y +z + w = 20

Q. If

x^{x} . y^{y} . z^{z} = x^{y} . y^{z} . z^{x}

such that x , y and z are positive integers greater than 1, then which of the following cannot be true for any of the possible value of x,y and z ?

Q. Q. How many triplets

(x, y, z)

satisfy the equation

x + y + z = 6

, where

x, y

and

z

are natural numbers?

The number of solutions (x, y, z) to the equation x - y - z = 25, where x, y, and z are positive integers such that $x \leq 40, y \leq 12$ , and $z \leq 12$ is

Q. If x + y + z = 1 and x,y,z are positive numbers such that

(1 - x) (1 - y) (1 - z) \geq k x y z

, then exact value of k is equal to

Join BYJU'S Learning Program