the no. of different necklaces formed by using 2n identical diamonds and 3 different jewels when exactly 2 jewels are always together -----

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First we will make necklace of 2n diamonds and as diamonds are identical, hencethere is only one way to do it.Now we have 3 different jewels when exactly 2 jewels are always together.Number of ways to select of two jewels which will be together=3C2Now we have 2n diamonds in the necklace so we have 2n vacant spaces  toput the two jewelswhich are together but as diamonds are identical hencewe will get same necklace wherever we put the two jewels. Hence there will beone way to put the two jewels.As one of the 2n places is occupied by occupied by 2 jewels hence last jewel, so will have2n-1 places to fit the last jewels.Number of necklaces=3C2×2n-1=3×22×1×2n-1=32n-1 =6n-3 Answer

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