Math 3113, section 4
Fall 1999
Solutions to the second test

(28 points) This problem is about license plates consisting
of 3 letters followed by 3 numbers,
for instance ABC 123.

Find the number of license plates which have a triple
letter.
(So, AAA 012 would be a particular instance.)
Solution: 26·1·1·10·10·10
= 26,000

Find the number of license plates which have a triple
number.
(So, XYZ 999 would be a particular instance.)
Solution: 26·26·26·10·1·1
= 175,760

Find the number of license plates which have a triple
letter anda triple number.
Solution: 26·1·1·10·1·1
= 260

Find the number of license plates which have a triple
letter ora triple number.
Solution: 26,000 + 175,760  260
= 201,500

(14 points) Expand by using the binomial theorem:
Solution: (x  2y)^{5}
= 2^{0}x^{5}y^{0}  5·2x^{4}y^{1}
+
· 2^{2}x^{3}y^{2}

· 2^{3}x^{2}y^{3}
+ 5·2^{4}x^{1}y^{4}
 2^{5}x^{0}y^{5},
so that (x  2y)^{5 }= x^{5}
 10x^{4}y + 40x^{3}y^{2}  80x^{2}y^{3}
+ 80xy^{4}  32y^{5}.

(16 points) A certain club has 20 members.

In how many ways is it possible to nominate 4 members
for the offices of President, Vice President, Secretary, and Treasurer
(in that order ... ) ?
Solution: 20·19·18·17
= 116,280

In how many ways is it possible to select a committee
consisting of 4 members of the club?
Solution:
=
= 4845

(14 points)
Two pairin a five card poker hand has two
of one kind, two of another kind, and the fifth card of a third kind, for
instance 22QQA. Determine the number of two pair hands.
Suggestion: Check to be sure that your
answer doesn't count anything more than once!
Method 1:·
{··
3 ···
3}· 44 = 2808 · 44 = 123,552
Method 2:·
{13 ··
12 · }·
44 = 2808 · 44 = 123,552
Remark: the "extra" factor of
outside the brackets is to eliminate duplicates because the two pairs in
a given hand could have been chosen in two different orders. For instance,
we should not count 22QQA and also QQ22A as being different hands!
Another way to see this. If we just do this calculation:
{
··
3} ·{ ·
·
3} · 44 ,
then we are counting the ways to do the following steps:
 choose one pair;
 choose another pair (with a different numerical value);
 choose a fifth card (with a still different numerical value).
So, if we have chosen two given pairs in a particular order,
we can get the same hand
by choosing the same two pairs in the the opposite order. To eliminate the
duplication, we have to multiply by 1/2.

(14 points) How many rearrangements of the letters of
the word ALABAMA are there?
Solution:·
3! = 210
(We first choose places for the four A's, and then
distribute the other 3 letters among the remaining places.)

(14 points) In how many ways can 10 identical coins
be distributed to 3 people?
Suggestion: This can be considered
as counting solutions of x_{1} + x_{2} + ... + x_{n}
= k.
In that case, one has to determine the correct values
for nand k.
Solution: We take n= 3 and k=
10. So, the number of solutions is ,
or
=
= 66.
For questions:
roberts@math.umn.edu
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