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16 different things are to be distributed among 3 persons A,B,C so that B gets 1 more than A and C get 2 more than B. In how many ways this can be done?

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1 Answer

Let A get $x$ things, then B gets $x+1$ things and C gets $x+3$ things.
$\therefore \:x+x+1+x+3=16$
$\Rightarrow\:x=4$
$i.e.,$ A gets 4 things, B gets 5 things and C gets 7 things.
$\therefore$ The required no. of ways = $^{16}C_4\times ^{12}C_5\times ^7C_7$ ways.
$=\large\frac{16!}{4!.5!.7!}$ ways.
answered Sep 4, 2013 by rvidyagovindarajan_1
 

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