Question

How many different necklaces can be formed with 6 white and 5 red beads?

• A. 18
• B. 24
• C. 21
• D. 27

Answer 1

The correct option is C 21

n = Total number of beads = 6 + 5 = 11
$P_1=6,P_2=5$
$\therefore$ Number of different necklaces
$=\frac{1(11-1)!}{26!5!}=\frac{10!}{2.6!5!}=\frac{\mathrm{10.9.8.7.6}!}{2.6!\mathrm{5.4.3.2.1}}=3\times 7=21.$