**Quote:**

This combinatorics problem can be approached in multiple ways depending on the story that one tells to describe it. One approach is to treat it as a pure linear permutation and then divide out duplicate orders to account for the circular nature of the arrangement. Another approach is to separate the problem into a combination to choose the people and then a circular arrangement to order them.

Taking the former approach, the problem is "yes order" and "no replacement," so it's a permutation. There are 7 choices for the first chair, 6 for the second, 5 for the third, 4 for the fourth, and 3 for the fifth. As a linear permutation, the answer would be 7∗6∗5∗4∗3

However, this permutation is circular, which means that we could have started placing children from any of the five seats. In other words, from a circular perspective, there is no real difference between the arrangement ABCDE and the arrangements BCDEA, CDEAB, DEABC, and EABCD. In fact, we have overcounted by this same factor of five in each of our orderings. So the real answer must be (7∗6∗5∗4∗3)/5 =7∗6∗4∗3=504

(Note that we could estimate, examine the units digits, and look at the answer choices to avoid having to put too much effort into this calculation; it's 42∗12, so it's greater than 400 and it ends in a 4) This is D.

**Quote:**

Another option is to treat the problem first as a combination. We have seven children, but only five will get seats. Determining which children get to sit is a "no order," "no replacement" proposition, so it's a combination, and we can use the formula

nCk=n!/(n−k)!∗k!

In this formula, n represents the number of items from which we will choose and k represents the number of picks that will be made. With n=7 and k=5, the formula gives

7C5=7!/2!∗5!=7∗6/2=21

Having chosen the five children who will sit, we turn to arranging them in a circle. We can use the rule that the first item is "free" in a circular arrangement. (It doesn't really matter where the first child sits, in terms of the relative positions of the children.) The arrangement can therefore be calculated as 1∗4∗3∗2∗1=24

Finally, since we must choose the children and arrange them, we want 21∗24=504

(again, estimate and/or look at units digits to save effort). The answer is D.

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