How many ways can seven children sit on a bench? How many of these arrangements are the two oldest always together?
7! = 5040
if the two oldest are one, then we only really have six
6! = 720
but one could be right and the other left or reverse so twice that many
*2 = 1440
To find the total number of ways seven children can sit on a bench, we can use the concept of permutations. In this case, since the order of the children matters, we need to calculate the number of permutations.
The total number of ways to arrange seven children is given by 7 factorial (7!). Factorial means multiplying a number by all the positive integers less than itself.
So, 7! = 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5,040.
Now, let's consider the second part of the question: how many arrangements have the two oldest children sitting together.
To solve this, we can consider the two oldest children as one pair. We can treat them as a single entity, which means there are six entities to arrange on the bench.
Now, the number of arrangements with the two oldest children always sitting together is given by 6 factorial (6!). We do not distinguish between the two oldest children since they are considered as a single entity.
So, 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720.
Hence, there are 720 arrangements where the two oldest children always sit together on the bench out of the total 5,040 possible arrangements.