ABC is a triangle. 4, 5, 6 points are marked on the sides AB, BC, CA respectively, the number of triangles having vertices on different sides is

Question

ABC is a triangle. 4, 5, 6 points are marked on the sides AB, BC, CA respectively, the number of triangles having vertices on different sides is (A) (4+ 5+ 6)! (B) (4-1)(5-1)(6-1) (C) 5! 4! 6! (D) 4 × 5 × 6

Tardigrade Admin · Accepted Answer

Required number of such triangles = 4C1 × 5C1 × 6C1 = 4 × 5 × 6

Q. $A BC$ is a triangle. $4, 5, 6$ points are marked on the sides $A B, BC, C A$ respectively, the number of triangles having vertices on different sides is

$(4 + 5 + 6)!$

$(4 - 1) (5 - 1) (6 - 1)$

$5! 4! 6!$

$4 \times 5 \times 6$

Required number of such triangles
$=^{4} C_{1} \times^{5} C_{1} \times^{6} C_{1}$
$= 4 \times 5 \times 6$

Q. ABC is a triangle. 4,5,6 points are marked on the sides AB,BC,CA respectively, the number of triangles having vertices on different sides is

(4+5+6)!

(4−1)(5−1)(6−1)

5!4!6!

4×5×6