In how many ways can 7 people be lined up in a row for a picture if two of them should have at least 2 people between them?
A hint would be to first
Consider drawing 7 slots for the people then organizing how many ways the line up can go...
Doris, _____, ______ Paul,_______, _______, ________ but then
_____, Doris, _______, _______, Paul,_____, ______ and then
_____, _____, Doris, ______, _______, Paul, _______ and then
_____, _____, _____, Doris, ______, _______, Paul but you could have Paul first and Doris second so you have to times by 2 for that arrangement.
So there are 4 ways Doris and Paul can be placed, then times by 2 for the Paul then Doris scenerio,
that is 4 x 2 x now the other 5 people are placed in the spots (so 5!)
4 x 2 x 5 x 4 x 3 x 2 x 1
now consider the word "at least" in the question...
Well, I can think of a few ways to arrange them. We could have the two special people sit on each other's shoulders, with the other five forming a human pyramid around them. That's one option. Or, we could have the two special people ride in on unicycles, with the other five forming a line behind them. That's another option. Oh, and we could also have the two special people dress up as clowns and have the other five people throw pies at them. That would certainly be a unique picture! So, in conclusion, there are definitely more ways than one to arrange these seven people in a row for a picture!
To solve this problem, we can treat the two people who should have at least 2 people between them as a single entity. Let's call this entity "AB".
Now, we have effectively reduced the problem to arranging 6 entities in a row: AB, 5 other people (C, D, E, F, G).
The 6 entities can be arranged in 6! (factorial) ways. However, since AB is considered as a single entity, we need to consider the arrangements of A & B within AB.
There are 2! ways to arrange A & B within AB.
Therefore, the total number of ways to arrange 7 people in a row with two people having at least 2 people between them is 6! * 2!.
Simplifying this expression:
6! = 6 x 5 x 4 x 3 x 2 x 1 = 720
2! = 2 x 1 = 2
Total ways = 720 x 2 = 1440
So, there are 1440 ways to line up the 7 people in a row for a picture if two of them should have at least 2 people between them.
To find the number of ways to line up 7 people in a row with the condition that two of them should have at least 2 people between them, we can use a combination of counting principles.
First, let's consider the two people who should have at least 2 people between them as a single entity. So effectively, we have 6 entities to arrange: 5 single individuals and the pair of two.
We can think of the pair as "locking" together, creating a single unit. Now, we have 6 entities to arrange in a row, where one of them is two people locked together.
The number of ways to arrange these 6 entities can be calculated using the factorial function. We have 6 entities, so the total number of arrangements is 6!.
However, the pair of two people within the locked entity can be arranged in 2! ways (since there are two people in it). Therefore, we need to account for this overcounting by dividing the total number of arrangements by 2!.
So, the number of ways to arrange the 7 people with the given condition is:
6! / 2! = (6 * 5 * 4 * 3 * 2 * 1) / (2 * 1) = 360 / 2 = 180 ways.
Therefore, there are 180 ways to line up 7 people in a row for a picture where two of them should have at least 2 people between them.