Given six line segments of lengths 2, 3, 4, 5 | vedclass

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(A) triangle can be formed if the sum of any two sides is greater than the third side.Total ways to select $3$ segments out of $6$ is $^6C_3 = \frac{6

Vedclass · Accepted Answer

$^6C_3 - 7$

Vedclass · Answer

(A) triangle can be formed if the sum of any two sides is greater than the third side.Total ways to select $3$ segments out of $6$ is $^6C_3 = \frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1} = 20$.We must subtract the combinations that do not form a triangle (where sum of two smaller sides $\le$ largest side).The sets of lengths that do $NOT$ form a triangle are:$(2, 3, 5), (2, 3, 6), (2, 3, 7), (2, 4, 6), (2, 4, 7), (2, 5, 7), (3, 4, 7)$.There are $7$ such combinations.Number of triangles = $20 - 7 = 13$.Looking at the options provided,the expression $^6C_3 - 7$ equals $20 - 7 = 13$.

Given six line segments of lengths $2, 3, 4, 5, 6, 7$ units,the number of triangles that can be formed by these lines is

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