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The number of ways in which 5 red beeds and 4 yellow beads of different sizes can be made out to form a necklace so that no two yellow beads come together is

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$4! \times^{5} p_{4}$

$\frac{6! \times^{4} p_{5}}{2}$

$\frac{4!^{5} p_{4}}{2}$

$6! \times^{4} p_{5}$

answer is A.

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Detailed Solution

5 red beds can be arranged in $\frac{(5 - 1)!}{2} = \frac{4!}{2}$ ways

In 5 gaps ,4 yellow beds can be arranged in $^{5} p_{4}$ ways

$= \frac{4!^{5} p_{4}}{2}$

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