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

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(A) The total number of ways to select $3$ line segments out of $6$ is given by $^6C_3 = \frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1} = 20$. $A$ tr

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$^6C_3 - 7$

Vedclass · Answer

(A) The total number of ways to select $3$ line segments out of $6$ is given by $^6C_3 = \frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1} = 20$. $A$ triangle can be formed if the sum of the lengths of any two sides is greater than the third side. Let the set of lengths be $S = \{2, 3, 4, 5, 6, 7\}$. The combinations that do $NOT$ form a triangle are those where the sum of the two smaller sides is less than or equal to the largest side: $1$. $(2, 3, 5)$ since $2+3=5$ $2$. $(2, 3, 6)$ since $2+3 $3$. $(2, 3, 7)$ since $2+3 $4$. $(2, 4, 6)$ since $2+4=6$ $5$. $(2, 4, 7)$ since $2+4 $6$. $(2, 5, 7)$ since $2+5=7$ $7$. $(3, 4, 7)$ since $3+4=7$ There are $7$ such combinations. Therefore,the number of triangles that can be formed is $20 - 7 = 13$. Looking at the options,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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