n balls each of weight w are weighed in pairs | vedclass

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(B) The number of possible pairs of $n$ objects is given by $\binom{n}{2} = \frac{n(n-1)}{2}$.Since each pair consists of $2$ balls,the total weight o

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$10$

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(B) The number of possible pairs of $n$ objects is given by $\binom{n}{2} = \frac{n(n-1)}{2}$.Since each pair consists of $2$ balls,the total weight 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 $\binom{n}{3} = \frac{n(n-1)(n-2)}{6}$.Since each triplet consists of $3$ balls,the total weight 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 $w$ are 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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