n balls each of weight w. When weighed in pai | vedclass

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(B) The number of possible pairs of $n$ objects is given by $^nC_2 = \frac{n(n-1)}{2}$.Each pair has a total weight of $2w$.Therefore,the sum of the w

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$10$

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(B) The number of possible pairs of $n$ objects is given by $^nC_2 = \frac{n(n-1)}{2}$.Each pair has a total weight of $2w$.Therefore,the sum of the weights of all pairs is $\frac{n(n-1)}{2} 	imes 2w = n(n-1)w = 120$  --- $(1)$.The number of possible triplets of $n$ objects is given by $^nC_3 = \frac{n(n-1)(n-2)}{6}$.Each triplet has a total weight of $3w$.Therefore,the sum of the weights of all triplets is $\frac{n(n-1)(n-2)}{6} 	imes 3w = \frac{n(n-1)(n-2)w}{2} = 480$  --- $(2)$.Dividing equation $(2)$ by equation $(1)$:$\frac{\frac{n(n-1)(n-2)w}{2}}{n(n-1)w} = \frac{480}{120}$$\frac{n-2}{2} = 4$$n-2 = 8$$n = 10$.

$n$ balls each of weight $w$. When weighed in pairs,the sum of the weights of all possible pairs is $120$. When they are weighed in triplets,the sum of the weights of all possible triplets is $480$. Then $n$ is:

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