There are $2$ blue, $4$ yellow and $3$ green identical marbles. In how many ways can they be arranged such that any two green marbles won't be adjacent?
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Let us evaluate in how many ways blue and yellow marbles can be arranged. Since the marbles are identical, we have that $$P = \dfrac{6!}{4!\cdot 2!}$$
There are also $7$ slots which we can fill in green marbles, $7$ slots out of $3$
$$7\times 6\times 5$$
Hence, we get
$$7\times 6\times 5 \times \dfrac{6!}{4!\cdot 2!}$$
Am I right?
Regards
When you select three slots out of seven you need to divide by $3!$ because the green marbles are identical. The number of ways to select them is $7 \choose 3$