The number of ways in which 20 different pearls of two colours can be set alternately on a necklace, there being 10 pearls of each colour, is

Question

TestHub · Accepted Answer

Ten pearls of one colour can be arranged in $\frac{1}{2} . (10 - 1)$ ! ways

The number of arrangements of 10 pearls of the other colour in 10 places between the pearls of the first colour $= 10! .2!$

∴ The required number of ways $= \frac{1}{2} \times 9! \times 10! .2!$

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