# Math/Stat 394 Homework 1 ```Math/Stat 394 Homework 1
1. (a) How many different 7-place license plates are possible if the first
2 places are for letters and the other 5 for number?
(b) Repeat part(a) under the assumption that no letter or number
can be repeated in a single license plate
(a) There are 26 choices for each of the two places for letters and
10 choices for each of the five places for letters. Thus by the
multiplication rule there are (26)2 (10)5 possibilities.
(b) There are 26 choices for the two first letter and 25 choices for
the second. There are 10 choices for the first letter, 9 choices for
the second, 8 for the third, 7 for the fourth and 6 for the fifth.
Thus by the multiplication rule there are (26)(25)(10)(9)(8)(7)(6)
possibilities.
4. John, Jim, Jay and Jack have formed a band consisting of 4 instruments. It each of the boys can play all 4 instruments, how many
different arrangements are possible? What if John and Jim can play
all 4 instruments, but Jay and Jack can each play only piano and
drums?
For the first part there are 4! = 24 ways to pair the boys to the instruments. Since they can all play all the instruments all 24 are possible.
For the second question we must have Jack and Jay playing the piano
and drums while John and Jim play the other two instruments. There
are two choices for who plays the piano and drums and two choices for
who plays the other two instruments. Thus there are four choices.
5. For years, telephone area codes in the United States and Canada consisted of a sequence of three digits. The first digit was an integer
between 2 and 9; the second digit was either 0 or 1; the third digit
was any integer between 1 and 9. How many are codes were possible?
How many area codes stating with a 4 were possible?
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There were 8 choices for the first digit, two choices for the second digit
and 9 choices for the third. Thus there were (8)(2)(9) possible area
codes. If the first digit is a 4 then there are two choices for the second
digit and 9 choices for the third. Thus there were (2)(9) possible area
codes that began with a 4.
7. (a) In how many ways can 3 boys and 3 girls sit in a row?
(b) In how many ways can 3 boys and 3 girls sit in a row if the boys
and the girls are each to sit together?
(c) In how many ways if only the boys must sit together?
(d) In how many ways if no two people of the same sex are allowed
to sit together?
(a) There are 6! = 720 possible arrangements. If there are no restrictions then all of them are allowed.
(b) Either the boys sit in the first three spots and the girls in the last
three or vice versa. For each choice of which spots the boys are
in there are 3! possible arrangements of the boys are 3! possible
arrangements of the girls. Thus there are 2(3!)(3!) = 72 possible
arrangements.
(c) In this case there are four choices for the places the boys can sit
(positions 1-3, 2-4, 3-5 or 4-6). For each choice of which spots the
boys are in there are 3! possible arrangements of the boys are 3!
possible arrangements of the girls. Thus there are 4(3!)(3!) = 144
possible arrangements.
(d) In this case there are two choices for the places the boys can sit
(positions 1,3 and 5 or positions 2,4 and 6). For each choice of
which spots the boys are in there are 3! possible arrangements
of the boys are 3! possible arrangements of the girls. Thus there
are 2(3!)(3!) = 72 possible arrangements.
10. In how many ways can 8 people be seated in a row if
(a) there are no restrictions on the seating arrangement;
(b) persons A and B must sit next to each other;
(c) there are 4 men and 4 women and no 2 men or women can sit
next to each other;
(d) there are 5 men and they must sit next to each other;
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(e) there are 4 married couples and each couple must sit together?
(a) If there are no restrictions then there are 8! possible arrangements.
(b) If persons A and B must sit together then there are 7 possible
places where they can sit (places 1 and 2, places 2 and 3, . . . ,
places 7 and 8). For each of those choices there are two choices
for which of the two positions A sits in. For each of these choices
the position of B is fixed and there are 6! possible arrangements
for the other 6 people. Thus there are (7)(2)(6!) possible arrangements.
(c) If there are 4 men and 4 women and no 2 men or women can sit
next to each other then there are 2 possible sets of places (1,3,5
and 7 or 2,4,6 and 8) where the women sit. For each of these
two choices there are 4! possible arrangements of the men and 4!
possible arrangements for the women. Thus there are (2)(4!)(4!)
possible arrangements with no two men sitting next to each other
and no two women sitting next to each other.
(d) If there are 5 men and 3 women and the men must sit next to
each other then there are 4 possible sets of places (1,2,3,4 and 5,
2,3,4,5 and 6, 3,4,5,6 and 7 or 4,5,6,7 and 8) where the men sit.
For each of these four choices there are 5! possible arrangements
of the men and 3! possible arrangements for the women. Thus
there are (4)(5!)(3!) possible arrangements with all 5 men sitting
next to each other.
(e) There are 4 choices for which couple sits in the first two spots.
For each of those choices there are three choices for the couple
in the third and fourth spots and 2 choices for the couple in the
fifth and sixth spots. For each choice of the order of the couples
there are two ways to seat each couple. Thus there are (4!)24
possible arrangements that have all four couples sitting next to
each other.
12. Five separate awards (best scholarship, best leadership qualities, and
so on) are to be presented to selected students from a class of 30. How
many different outcomes are possible if
(a) a student can receive any number of awards;
(b) each student can receive at most 1 award?
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(a) If there are no restrictions on how many awards a student can
receive then there are 30 choices for each of the five awards. Thus
there are 305 choices for who wins the five awards.
(b) If each student can receive only one award then there are 30
choices for the first award, 29 choices for the second award, 28
for the third, 27 for the fourth and 26 for the fifth. Thus there
are (30)(29)(28)(27)(26) choices for who wins the five awards.
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