જો દરેક દડો w વજન ધરાવતો હોય તેવા n દડાઓ | vedclass

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Question

${\,^n}{{\rm{C}}_2} = \frac{{{\rm{n}}({\rm{n}} - 1)}}{2} = $ number of possible pairs

of $n$ objects.

Total weight of $\frac{\mathrm{n}(\mathrm{

Vedclass · Accepted Answer

$10$

Vedclass · Answer

${\,^n}{{m{C}}_2} = \frac{{{m{n}}({m{n}} - 1)}}{2} = $ number of possible pairs

of $n$ objects.

Total weight of $\frac{\mathrm{n}(\mathrm{n}-1)}{2}$ pairs are

$\frac{\mathrm{n}(\mathrm{n}-1)}{2} 	imes 2 	imes \mathrm{w}=\mathrm{n}(\mathrm{n}-1) \mathrm{w}=120 $           .........$(1)$

similarly total weight of all triplets $=480$

$\Rightarrow \frac{n(n-1)(n-2) w}{2}=480 $            ......$(2)$

dividing $( 2)$ by $(1),$ we get $\frac{n-2}{2}=4 \Rightarrow n=10$

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