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. $6n$
• C. $6n-6$
• D. None of these

Answer 1

Answer: (A)

$6n-3$

first lets make a diamond necklace with the $2n$ diamonds

Since all diamonds are identical there is only one possible way of doing this

Consider case of $3$ different jewels of which $2$ of them are always together

Number of ways to select $2$ of the $3$ jewels ${=}^3{C}_2$

For $2n$ diamonds we have $2n$ vacant spaces where these two jewels can be placed

As diamonds are identical wherever we place these two gems we will get same necklace. Hence there is only way

If one of $2n$ vacant spaces is filled with gem we have left with $2n-1$ vacant spaces

total number of possible necklaces ${=}^3{C}_2\times 2n-1$

$=\frac{3\times 2}{2}\times (2n-1)$

$=6n-3$