Seven children â€” A, B, C, D, E, F, and G â€” are going to sit in seven chairs in a row. The children C & F have to sit next to each other, and the others can sit in any order in any remaining chairs. How many possible configurations are there for the children? Choices.
A) 600  B) 720  C) 1440  D) 4320  E) 4800
There are six different pairs of â€œnext to each otherâ€ chairs. For each pair, children C &amp; F could be in either order, so thatâ€™s  6*2= 
<gadget id="calculator"> 6*2</gadget>
<output>12</output>
12 possibilities for these two.   Now, consider the other five children. For any configuration of C &amp; F, the remaining five children could be seated in any order among the five remaining seats. Five items in any order â€” thatâ€™s a permutation of the 5 items â€”- 5P5= 5! = 120. For any single configuration of C &amp; F, there are 120 ways that the other children could be seated in the remaining seats.   Finally, weâ€™ll combine with the Fundamental Counting Principle. We have 12 ways for the first two, and 120 ways for the remaining five. Thatâ€™s a total number of configurations of  12*120= 
<gadget id="calculator"> 12*120</gadget>
<output>1_440</output>
1440. Answer= 1440
<result>C</result>
Seven children â€” A, B, C, D, E, F, and G â€” are going to sit in seven chairs in a row. The children C & F have to sit next to each other, and the others can sit in any order in any remaining chairs. How many possible configurations are there for the children?