(B) Step $1$: Select $4$ glass beads from $6$ available: $^6C_4 = 15$ ways.Step $2$: Select $4$ metal beads from $5$ available: $^5C_4 = 5$ ways.Step

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$^6C_4 \times ^5C_4 \times \frac{7!}{2!}$

(B) Step $1$: Select $4$ glass beads from $6$ available: $^6C_4 = 15$ ways.Step $2$: Select $4$ metal beads from $5$ available: $^5C_4 = 5$ ways.Step $3$: Total beads selected = $4 + 4 = 8$ beads.Step $4$: The number of ways to arrange $n$ distinct beads in a necklace is $\frac{(n-1)!}{2}$.Step $5$: For $8$ beads,the number of arrangements is $\frac{(8-1)!}{2} = \frac{7!}{2!}$.Step $6$: Total number of necklaces = $^6C_4 \times ^5C_4 \times \frac{7!}{2!}$.

Class 11 Mathematics Permutation and Combination Geometrical problems

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How many necklaces can be formed by selecting $4$ beads from $6$ different colored glass beads and $4$ beads from $5$ different colored metal beads?

A
$^6P_4 \times ^5P_4 \times \frac{7!}{2!}$
B
$^6C_4 \times ^5C_4 \times \frac{7!}{2!}$
C
$^6C_4 \times ^5C_4 \times \frac{8!}{2!}$
D
$^6C_4 \times ^5C_4 \times 7!$

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How many necklaces can be formed by selecting $4$ beads from $6$ different colored glass beads and $4$ beads from $5$ different colored metal beads?

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Let $n \geq 2$ be an integer. Take $n$ distinct points on a circle and join each pair of points by a line segment. Colour the line segment joining every pair of adjacent points blue and the rest red. If the number of red and blue line segments are equal,then the value of $n$ is

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