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The no. of ways in which 20 different pearls of two colours can be set alternatively on a necklace, there being 10 pearls of each colour, is ?

$\begin{array}{1 1} 9!.10! \\ 5.(9!)^2 \\ (9!)^2 \\ 5. (10 !)^2 \end{array}$

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  • No.of arrangement of n things of same colour can be arranged in a circle in $\large\frac{(n-1)!}{2}$ ways.
Pearls of one colour can be arranged in $\large\frac{(10-1)!}{2}=\frac{9!}{2}$
10 pearls of the other colour are arranged in 10 places between the pearls of $1^{st}$ colour.
in $10!$ ways.
$\therefore$ the required no. of arrangements = $\large\frac{9!}{2}$$\times 10!=5\times (9!)^2$
answered Sep 5, 2013 by rvidyagovindarajan_1

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