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How many necklaces of 12 beads can be ma...

How many necklaces of 12 beads can be made from 18 beads of various colours?

A

`(118xx13!)/(2)`

B

`(110xx14!)/(2)`

C

`(119xx13!)/(2)`

D

`(110xx12!)/(2)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of how many necklaces of 12 beads can be made from 18 beads of various colors, we will follow these steps: ### Step 1: Selecting the Beads First, we need to select 12 beads from the 18 available beads. The number of ways to choose 12 beads from 18 can be calculated using the combination formula: \[ \text{Number of ways to choose 12 beads from 18} = \binom{18}{12} = \frac{18!}{12! \times (18-12)!} = \frac{18!}{12! \times 6!} \] ### Step 2: Arranging the Beads Next, we need to arrange the 12 selected beads into a necklace. When arranging beads in a circular form, the number of arrangements is given by \((N-1)!\) where \(N\) is the number of beads. However, since necklaces can be flipped (i.e., they can be viewed from both sides), we need to divide by 2: \[ \text{Number of arrangements of 12 beads in a necklace} = \frac{(12-1)!}{2} = \frac{11!}{2} \] ### Step 3: Total Number of Necklaces Now, we can find the total number of necklaces by multiplying the number of ways to choose the beads by the number of arrangements: \[ \text{Total number of necklaces} = \binom{18}{12} \times \frac{11!}{2} \] Substituting the combination formula into this equation gives: \[ \text{Total number of necklaces} = \frac{18!}{12! \times 6!} \times \frac{11!}{2} \] ### Step 4: Simplifying the Expression Now, we can simplify the expression: 1. The \(11!\) in the numerator cancels with part of the \(12!\) in the denominator: \[ \frac{18!}{12! \times 6!} \times \frac{11!}{2} = \frac{18!}{12 \times 11! \times 6!} \times \frac{11!}{2} = \frac{18!}{12 \times 6! \times 2} \] 2. This can be further simplified: \[ = \frac{18!}{2 \times 12 \times 6!} \] ### Final Answer Thus, the total number of necklaces that can be made is: \[ \text{Total number of necklaces} = \frac{18!}{2 \times 12 \times 6!} \]
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