# Combination and permutation (round table)

Given that 4 children and 6 adults are seated at a round table and there are 4 cupcakes a)How many ways can they be seated if at most 3 child sits together?

b)What's the probability that each child received a cupcake?

For a I did total number of ways they can be arranged - number of ways 4 children sit together so is it (10-1)! - (6P6 × 4P4)?

The 10 people can be arranged in a total of $(10-1)!$ ways.
For the children to all sit together you can think about there only being 7 items now - the six adults and the group of children. These can be arranged in $(7-1)!$ ways. Then within the children there are different orders they could sit in. Four children means $4!$ ways.
So the answer to a) is $9!-6!\cdot4!$ which is what you had.
$$\frac{1}{10\choose4}=\frac{1}{\frac{10\cdot9\cdot8\cdot7}{4\cdot3\cdot2\cdot1}}=\frac{1}{210}$$