Question

The number of different necklaces formed by using $2n$ identical diamonds and $3$ different jewels when exactly two jewels are always together is ?

• A. $6n-3$
• B. $6$ $n$
• C. $6n-6$
• D. none of thes

Answer 1

The correct option is A $6n-3$

for a diamond necklace, we need with $2n$ diamonds

Now consider case of $3$ diff. jewels of which $2$ of them are always

No. of ways $={}^3{C}_2$

Now Since we have $2n$ diamonds, we have $2n$ vacant spaces

But since diamonds are identical wherever we place them

ways $=1$

Now for $2n$ vacant space is occupied by two jewels

Hence, $(2n-1)$ vacant space left

No. of possible necklace= ${}^3{C}_2(2n-1)$

$=6n-3$