Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row.
Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Child A has to sit next to both BG, with these two children immediately adjacent to here on either side. The other four children can sit in any order in any of the remaining seats. How many possible configurations Y are there for the children?
Answer/Solution
240
Steps/Work
MAGOOSHOFFICIAL SOLUTION:
First, we will consider the restricted elements — children ABG have to be in three seats in a row. How many “three in a row” seats are there in a row of seven seats?
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There are five different “three in a row” locations for these three children. Now, for any given triplet of seats, we know A has to be in the middle, so the children could be seated B-A-G or G-A-B — just those two orders. This means the total number of configurations for these three children is 5*2 = 10.
Now, consider the non-restricted elements, the other four. Once ABG are seated, the remaining four children can be seated in any order among the remaining four seats — that’s a permutation of the 4 items —- 4P4 = 4! = 24. For any single configuration of ABG, there are 24 ways that the other children could be seated in the remaining seats.
Finally, we’ll combine with the Fundamental Counting Principle. We have 10 ways for the first three, and 24 ways for the remaining four. That’s a total number of configurations Y of 24*10 = 240.
Answer = A
First, we will consider the restricted elements — children ABG have to be in three seats in a row. How many “three in a row” seats are there in a row of seven seats?
X X X _ _ _ _
_ X X X _ _ _
_ _ X X X _ _
_ _ _ X X X _
_ _ _ _ X X X
There are five different “three in a row” locations for these three children. Now, for any given triplet of seats, we know A has to be in the middle, so the children could be seated B-A-G or G-A-B — just those two orders. This means the total number of configurations for these three children is 5*2 = 10.
Now, consider the non-restricted elements, the other four. Once ABG are seated, the remaining four children can be seated in any order among the remaining four seats — that’s a permutation of the 4 items —- 4P4 = 4! = 24. For any single configuration of ABG, there are 24 ways that the other children could be seated in the remaining seats.
Finally, we’ll combine with the Fundamental Counting Principle. We have 10 ways for the first three, and 24 ways for the remaining four. That’s a total number of configurations Y of 24*10 = 240.
Answer = A