Question 1

Among a group of 3 cats (Tabby, Leo, and Garfield) and 4 dogs
(Snoopy, Odie, Fido, and Bowser) we wish to arrange a
photograph of 5 of them.
How many ways are there to arrange
5 of the animals in a photograph under the following
conditions (each set of conditions is independent of the others):
1) No conditions; any 5 animals may appear in any order.
2) There must be exactly 3 dogs in the photograph.
3) A cat must be in either the leftmost position or the
rightmost position.
Hint:
Use inclusionexclusion.
Answer to Question 1:

(1)
Worth 7 points.
There are 7 choices for the leftmost
position, 6 remaining choices
for the next position, 5 for
the next, 4 for the next, and 3 for the rightmost position.
By the
product rule, we have P(7,5) = 7*6*5*4*3 = 2520
arrangements.
Little partial credit was
given for errors
here.
Three points were awarded for calculating C(7,5)
instead of P(7,5).
(2)
Worth 6 points.
There are C(4,3) = C(4,1) = 4 ways
to choose 3 dogs for the
picture.
Then there are 5*4*3 ways
to choose positions for them in the picture.
Then we
have
to choose 2 cats for the other 2 spots in the picture.
There are C(3,2) = 3 ways to do
this, and then 2 ways to
arrange the cats in the 2 vacant spots left in the picture.
Multiplying all of these choices together, we get
4*(5*4*3)*3*2 = 1440.
Partial credit was
given only in
very special cases, because there is little work to show other
than writing
down the factors in the multiplication.
(3) Worth 7 points.
We use inclusionexclusion.
Let S be the set of all arrangements
with a cat in the
leftmost position, and let T be the set of all arrangements
with a cat in the
rightmost position.
We want to find
S U T.
By inclusionexclusion,
S U T = S + T  S intersect T.
Now we find S.
There are 3 choices for a cat in the
leftmost position.
The
remaining spots of the photo can be
filled out without restriction.
So there are 6*5*4*3
ways
do accomplish this.
Hence, S = 3*6*5*4*3 =
1080.
By symmetry, T = 1080.
Finally, we must calculate
S intersect T, the number of arrangements with cats
in the left
and rightmost positions.
There are 3 ways
to choose a cat for the leftmost position.
Then
2 choices remain for a cat in the rightmost position.
The middle 3 spots can then be filled
out without restriction
in 5*4*3 ways.
So we get a total of
S intersect T = 3*2*5*4*3 = 360.
Hence, S U T = 1080 + 1080  360 = 1800.
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 Spring '08
 STRAIN
 Prime number, Chinese Remainder

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