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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)?

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You have taken the right approach to consider the number of ways all the children can sit together and subtract this from the total arrangements.

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.

For part b) we have to assume that the cupcakes are handed out randomly as it is not stated. (If there was real life then all the children would sit in front of the cupcakes.) Also the arrangement of the children is irrelevant. To pick all four children can be done in exactly 1 way so the probability is:

$$\frac{1}{10\choose4}=\frac{1}{\frac{10\cdot9\cdot8\cdot7}{4\cdot3\cdot2\cdot1}}=\frac{1}{210}$$

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