Question

seven people (a, b, c, d, e, f, g) are lining up to enter a museum. 1. how many ways are there to line up the seven people if b is somewhere before g, although not necessarily immediately before? 2. how many ways are there to line up the seven people if e, f and g are next to each other? 3. how many ways are there to line up the seven people if c must be next to e but d and f can not be next to each other?

Answer 1

In three different scenarios, B must be before G in 240 ways, E, F, and G together can be arranged in 720 ways, and with the restrictions on C, D, and F, there are 240 valid arrangements.

Step-by-step explanation:

To solve such permutation problems, we need to understand the restrictions given and apply combinatorial rules accordingly.

1. Ordering with B before G: When B must come before G in line, we consider them as a single unit but with two different internal arrangements (B first or G first), essentially blocking out one of the slots for G. The remaining five people can be arranged in any order, giving us a total of 5! arrangements. Because B and G in the unit can also be arranged in two ways (B then G, or G then B), we multiply the number of arrangements by 2, which yields 2 × 5! = 240 ways.
2. E, F, and G grouping: If E, F, and G must be next to each other, we treat them as one single unit. This single unit and the other four people can be arranged in 5! ways. Since E, F, and G can be arranged in 3! ways within their own group, we multiply these to get 5! × 3! = 720 ways.
3. Restrictions with C, D, and F: C must be next to E, and D cannot be next to F. Treating C and E as one unit (but with two possible internal arrangements), and ignoring the restriction about D and F initially, we'd have 5! arrangements for the five units (C-E, A, B, D, F). However, we also need to subtract the arrangements where D and F are together. There are 4 initial configurations (DF, FD, CD-E, A, B; D, CD-E, A, B, DF, etc.) with 4! ways to arrange each. Inside each configuration, C-E and DF can have 2! internal arrangements. We subtract the unwanted arrangements from the total: 5! × 2 - 4! × 2 × 2 = 240 ways.

Answer 2

The number of ways to line up seven people with b before g is 1440. When e, f, and g must be next to each other, there are 720 ways. For c to be next to e, and d not next to f, there are 480 ways.

Step-by-step explanation:

The questions presented involve counting the number of permutations of seven people (a, b, c, d, e, f, g) under different constraints and belong to the subject of mathematics, specifically permutation and combination concepts that are commonly studied in high school.

Answer 1

For the first question, to determine the number of ways to line up seven people with the constraint that b must be before g, we first consider the pair (b, g) as a single entity. There are six entities to arrange (a, c, d, e, f, and the pair (b, g)), leading to 6! permutations. Since b and g can be swapped, we multiply the result by 2, giving us 2*6! = 1440 ways.

Answer 2

For the second question, regarding the lineup of seven people with e, f, and g being next to each other, we first consider the trio (e, f, g) as one entity. This gives us five entities to arrange (a, b, c, d, and the trio (e, f, g)). We have 5! arrangements for these entities. However, within the trio, e, f, and g can be arranged in 3! ways. This gives us a total of 5! * 3! = 120 * 6 = 720 ways.

Answer 3

Finally, the third question asks for the number of ways to lineup seven people with the constraints that c must be next to e, and d and f cannot be next to each other. Treating c and e as a single entity, we have six entities to arrange. There are 6! ways to do this. However, since d and f cannot be next to each other, we need to calculate the total permutations without this restriction (6!) and subtract those where d and f are together. Treating d and f as a single entity gives us 5! ways to arrange everything with 2! additional ways to arrange d and f within this single entity, or 5! * 2! = 240 ways. Thus, the answer is 6! - 240. For the total number of permutations (6! = 720) and the subtracted permutations (240), the final answer is 720 - 240 = 480 ways.