Video Transcript
The names of four students are each
written on a piece of paper, which are then placed in a hat. If two names are randomly selected
from the hat, determine the number of all two-student selections that are
possible.
We’re looking to work out the
number of ways of choosing two names from a total of four. And so we’re going to begin by
deciding whether we’re interested in calculating the number of combinations or the
number of permutations. And this all comes down to whether
order matters. Specifically, if we’re choosing
from a selection of items and we say that order doesn’t matter, then we say that
that is a combination. If, however, order does matter,
then we have a permutation.
Let’s think about what happens when
we choose the two names for a hat. Suppose the two names we pull out
of the hat are Ali and Ben. It does not matter whether we
choose Ali first and then Ben or we choose Ben first and then Ali. This means that we’re interested in
the number of combinations. Order does not matter here. So let’s remind ourselves of the
formula we use. To calculate the number of
combinations of choosing 𝑟 items from a collection of 𝑛, we find 𝑛 choose 𝑟,
which is equal to 𝑛𝑃𝑟 over 𝑟 factorial.
When calculating from the
beginning, it can make sense to use the amended formula 𝑛 factorial over 𝑟
factorial times 𝑛 minus 𝑟 factorial. Now, 𝑛 is the total number of
items within the collection. Well, here this is the names of the
students. So there are four students and 𝑛
is equal to four. We’re choosing two names out of the
hat. So we’re going to let 𝑟 be equal
to two. We can therefore say that the total
number of ways of choosing these names is given by four choose two.
Substituting into the formula, and
we see that that is equal to four factorial over two factorial times four minus two
factorial. That simplifies to four factorial
over two factorial times two factorial. And then since two factorial is two
times one, it’s two, and four factorial can be equivalently written as four times
three times two factorial, we see we can divide through by a common factor of two
factorial and another factor of two. This means four choose two is equal
to two times three over one. And that’s simply six. And so there are a total of six
two-student selections that are possible.
