Hi michned,
I hope you find the following answer satisfactory. If anything is
unclear or if you want more explanation for any part, please request a
clarification before rating and I wll do my best to clear up any
confusion.
First a couple of definitions: nPk is the permutation of k elements
from a set of n--that is, the number of ways we can pick k things from
a group of n in such a way that order matters. In terms of factorials,
nPk = n!/(n-k)!
nCk is the combination of k elements from a group of n, that is, the
number of ways we can choose k elements from a group of n in such a
way that order doesn't matter. In factorials, nCk = n!/((n-k)! * k!)
For more information about combination, permutation, and factorials, see
< http://www.themathpage.com/aPreCalc/permutations-combinations.htm >
We'll start with part A. We know two of the people in our group are
Kristen and Karen, so just have them stand over to the side. They're
already in our picture group.
Now there are eight people left. We need to choose a further four
people from those eight to go in the picture taking group. It doesn't
matter which order we pick the people to photograph--they will be
ordered later when we arrange the photo. This means we are in an nCk
situation. There are 8C4 ways of choosing four people from a group of
eight. So that means that there are 8C4 possible groups that we might
be taking pictures of. 8C4 = 8!/((8-4)! * 4!) = 70 possible groups to
photograph.
Now each of those 70 groups has six people, counting Karen and
Kristen. So we need to figure out how many different orders those six
people can be put in, say, from left to right. The number of ways to
arrange all n members of a group in such a way that order matters is
n!. So that means if we are arranging all six people in some
particular order, there are 6! = 720 ways to do it.
Now if there are 70 possible groups, and 720 ways to order each group,
that means there are 70*720=50,400 different possible photograph
arrangements that must include Kristen and Karen.
The next part of the question is pretty interesting, because it
requires us to use a very clever trick. Note that there are still 70
possible groups for us to photograph. The question is now, what if
Kristen and Karen have to be next to each other? Well, let's do this,
then. Pretend that instead of Kristen and Karen being separate, we tie
them together with a bit of rope, as though they were in a
three-legged race. Now wherever Kristen goes, Karen has to go, and
vice versa. So that means, for the purposes of arranging the people in
the photograph, we can treat KristenKaren as a single person. That
means that instead of arranging 6 people, we are now effectively
arranging 5--since Kristen and Karen kind of count as a single person.
So the number of ways we can arrange the "5" people in some order is
5!=120. So the number of possible arrangements of the group that has
the KristenKaren monster is 70*120=8400 ways.
Now here's the thing: We originally might have tied Kristen on the
left and Karen on the right, or vice versa. Let's say our arrangement
was, say, something like:
C D B KK A
with Kristen and Karen tied together. This actually counts as two
different arrangements, since we could have bound the pair so that
either Kristen or Karen is on the left. So that number 8400 needs to
be doubled--each arrangement involving the KristenKaren monster
actually counts for two possible arrangements: one with Kristen on
left and Karen on right, and one with Karen on left and Kristen on
right. So the total number of ways to arrange the photo with Kristen
and Karen next to each other is 8400*2 = 16,800 possible arrangements.
I hope this explanation makes sense to you. If you are dissatisfied,
or if you need further explanation in any aspect of this problem,
please request a clarificaiton and I will do my best to make things
clearer.
Thanks,
smudgy. |
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