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6.Permutation and Combination
normal
$n$ balls each of weight $w$ when weighted in pairs the sum of the weights of all the possible pairs is $120$ when they are weighed in triplets the sum of the weights comes out to be $480$ for all possible triplets, then $n$ is
A
$5$
B
$10$
C
$15$
D
$20$
Solution
${\,^n}{{\rm{C}}_2} = \frac{{{\rm{n}}({\rm{n}} – 1)}}{2} = $ number of possible pairs
of $n$ objects.
Total weight of $\frac{\mathrm{n}(\mathrm{n}-1)}{2}$ pairs are
$\frac{\mathrm{n}(\mathrm{n}-1)}{2} \times 2 \times \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 $\frac{n-2}{2}=4 \Rightarrow n=10$
Standard 11
Mathematics
