The number of points, having both co-ordinates as integers, that lie in the interior of the triangle with vertices (0, 0), (0, 41) and (41, 0) is

Question

The number of points, having both co-ordinates as integers, that lie in the interior of the triangle with vertices (0, 0), (0, 41) and (41, 0) is (A) 901 (B) 861 (C) 820 (D) 780

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/660178615cfd25636edfdc6688acb39d-.png" /> x+y<41 1 to each x+y < 39 ⇒ x+y ≤ 38 ⇒ x+y+z=38 38+3-1 C 3-1= 40 C 2=(40 × 39/2)=780 .

Q. The number of points, having both co-ordinates as integers, that lie in the interior of the triangle with vertices $(0, 0), (0, 41)$ and $(41, 0)$ is

901

861

820

780

$x + y < 41$
$1$ to each
$x + y < 39$
$\Rightarrow x + y \leq 38$
$\Rightarrow x + y + z = 38$
$^{38 + 3 - 1} C_{3 - 1} =^{40} C_{2} = \frac{40 \times 39}{2} = 780.$

Q. The number of points, having both co-ordinates as integers, that lie in the interior of the triangle with vertices (0,0),(0,41) and (41,0) is

901

861

820

780

x+y<41 1 to each x+y<39 ⇒x+y≤38 ⇒x+y+z=38 38+3−1C3−1​=40C2​=240×39​=780.

Q. The number of points, having both co-ordinates as integers, that lie in the interior of the triangle with vertices $(0, 0), (0, 41)$ and $(41, 0)$ is

$x + y < 41$
$1$ to each
$x + y < 39$
$\Rightarrow x + y \leq 38$
$\Rightarrow x + y + z = 38$
$^{38 + 3 - 1} C_{3 - 1} =^{40} C_{2} = \frac{40 \times 39}{2} = 780.$