Problem Set 3: Solutions - Personal Pages - Denison University

fys102: Mathematical Game Theory
Problem Set 3: Solutions
1. How many ways can you arrange the word gate to form new words. (Here
will will always count ”nonsense” words too if they form a different arrangement).
4! = 24
2. How many anagrams are there for serial
6! = 720
3. How many anagrams are there for tote
4!
2! = 12
4. How many anagrams are there for giggle
6!
3! = 120
5. How many anagrams are there for banana
6!
3!·2! = 60
6. How many anagrams are there for mississippi
= 34, 650
11!
4!·4!·2!
7. You have five gumballs (two grape, two orange, and one cherry) to give to
five different children. In how many different ways can you distribute the
gumballs so that each child receives one
5!
2!·2! = 30
8. Girl Scout Troop #4 is rebuilding a local hiking trail. They have ten young
ladies to share the labor. They need four shovelers, four pick axers, and two
stone movers. In how many ways can they assign tasks to the ten girl scouts
10!
4!·4!·2!
= 3, 150
9. How many anagrams of attic begin with a vowel
First consider anagrams that begin with ”a”, there are 4!
2!
= 12 of them.
There are also 12 anagrams that begin with ”i”. So there are 24 anagrams
that begin with a vowel.
10. In computer science, a bit is a binary digit. Each bit can either be ”1” or
”0”. A collection of eight bits make a byte. How many different bytes are
possible
Easy way: each bit location has two choices: 2 8 = 256. Harder way: break
this down by cases depending upon how many ”1”s are present and then treat
1

each as an anagram.
# 1’s # anagrams
Zero 1’s 1 way
8!
One 1
7! = 8 ways
8!
Two 1’s
6!·2!
= 28 ways
8!
Three 1’s
5!·3!
= 56 ways
8!
Four 1’s
4!·4!
= 70 ways
8!
Five 1’s
3!·5!
= 56 ways
8!
Six 1’s
2!·6!
= 28 ways
8!
Seven 1’s
7! = 8 ways
Eight 1’s 1 way
total (sum) 256
11. A byte has even parity if the number of ”1”s is an even number. Otherwise a
byte has odd parity. How many bytes have even parity
Easy way: there must be exactly half of the bytes with even parity because
you can match each byte up with one that has odd parity by switching all
the 1’s and 0’s. So there are 258
2
= 128 bytes with even partiy. Harder way:
refer to the above table looking only at those cases where the number of 1’s
is even. Add them up to get 128.
12. Twenty-three people are attending Jane and Mark’s rehearsal dinner. If there
are eight steak entrees and fifteen chicken entrees, how many ways are there
to feed the dinner guests
23!
8!·15!
= 490, 314
13. In the above question, if there are two non-beef eaters at Jane and Mark’s
rehearsal dinner, then how many ways are there to distribute the meals
= 203, 490
21!
8!·13!
14. See the map of Big City below. You are to take a taxi from the corner of
Second Avenue and Third Street to the corner of Eighth Avenue and Seventh
Street. Taxis in Big City charge by total distance (not time), so the taxi must
take one of the shortest routes (hint: it is not allowed to ever drive West or
North for this particular trip). How many different routes are possible
Note that the cab must drive exactly 6 blocks east and 4 blocks south. The
various routes are formed by exchanging the order of the blocks. This is
exactly like an anagram problem in which there are ten letters (four are ”S”
and six or ”E”). We know there are 10!
4!·6!
= 210 ways for the cab to make this
trip.
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First Avenue
Second Avenue
Third Avenue
Fourth Avenue
Fifth Avenue
Sixth Avenue
Seventh Avenue
S
Eighth Avenue
First Street
Second Street
Third Street
Fourth Street
Fifth Street
Sixth Street
E Seventh Street
Eighth Street
Figure 1: Map of Big City
15. Referring to the previous question, assume there is an accident that has
closed the intersection of Fifth Avenue and Fourth Street (you cannot pass
through here). How many trips are possible with the cab now
First we count all the ways the cab could make the trip if it went through the
accident intersection (these are the illegal trips). We must take into account
both the ways to get from the start to the accident and then also from the
accident to the end. There are 4!
3! · 6!
3!·3!
= 80 illegal trips leaving us 210-80 =
130 legal trips that do not go through this intersection.
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