Download Homework Week 2 Solutions 1. How many license plates involving

Homework Week 2 Solutions
1. How many license plates involving one, two, or three letters and one, two, or three digits are there if
the letters must appear in a consecutive grouping?
Since the letters have to appear in a consecutive grouping, we can think of them as one big letter.
If there are 3 digits, then there are 4 spots for where the big letter goes (no matter if there are one
two or three letters). Thus, the answer is:
(no. of license plates with one digit) + (no. of license plates with two digits) + (no. of license plates
with three digits) =
2 · 10 · (26 + 262 + 263 ) + 3 · 102 · (26 + 262 + 263 ) + 4 · 103 · (26 + 262 + 263 ) = 4320 · 18278 = 78960960
2. How many different five-letter sequences can be made using the letters A, B, C, D with repetition
such that the sequence does not include the word BAD, i.e., sequences such as ABADD are excluded?
Number of total words - Number of “BAD” words:
Number of total words is 45 = 1024.
Number of BAD words look like BADxx, xBADx, xxBAD (where x is any letter). There are 16 choices
for each time of BAD word; for a total of 48 BAD words. Thus, there are 976 good sequences.
3. Consider the identity
µ ¶µ ¶ µ ¶µ
¶
n
m
n
n−k
=
.
m
k
k
m−k
(a) Prove this identity using an “algebraic” proof.
(b)
LHS =
m!
n!
n!
·
=
m!(n − m)! k!(m − k)!
k!(n − m)!(m − k)!
RHS =
(n − k)!
n!
n!
·
=
= LHS
k!(n − k)! (m − k)!(n − m)!
k!(m − k)!(n − k)!
∗
Prove this identity using a “combinatorial” proof.
Suppose you have n objects and you want to select m − k “sort of special” items and k ‘very
special” items. To do this, you can either (a) choose m items to be considered “special.” From
those m, you choose k of them to be “very special.” The rest (m − k) are just “sort of special.”
¡ n ¢¡m¢
The number of ways to choose these items in that manner is the LHS of the equality: m
k
Option (b) is to choose the k “very special” items first. Then you can choose from the remaining
n − k items, the ¡m¢¡
− k ”sort
¢ of special” items. The number of ways to do option (b) is the RHS
n−k
of the equality: nk m−k
.
4. How many 10-letter “words” are there using 5 different vowels and 5 different consonants?
Choose which places the vowels occupy among the 10 spots:
¡10¢
5 .
Now choose which vowels are used and where they go (well, you’ll be using all 5 of them): 5! ways
to do this.
Now choose which consonants are used and where they go: P (26, 5) = 26 · 25 · 24 · 23 · 22
¡ ¢
The grand total is: 10
5 · 5! · 26 · 25 · 24 · 23 · 22.
HW pp. 60 − 68 :
#5b The number of zeros in 1000! is just the number of 5’s that appear in the product.
Every 5th number has a 5 in its prime factorization, so that leads to 200 zeroes.
Every 25th number has another 5 in its prime factorization, so that leads to 1000 / 25 = 40 more zeroes.
Every 125th number (53 ) has another 5 in its prime factorization, so that leads to 1000 / 125 = 8 more
zeroes.
Every 625th number (54 ) has another 5 in its prime factorization, so that leads to b1000/625c = 1 more
zero.
The total number of zeroes is 249.
#6
If the number has 4 digits and the first digit is a 5, then we have 1 · 4 · 6 · 5 choices. This is because the
2nd digit cannot be a 0, 1, 2, 3, 5, or 7.
If the number has 4 digits and the first digit is not a 5, then we have 3 · 7 · 6 · 5 choices. Here, we don’t
have as many restrictions on the 2nd digit. It cannot be a 2, 7, or the 1st digit.
If the number has 5 digits, then we have 7 · 7 · 6 · 5 · 4 choices.
If the number has 6 digits, then we have 7 · 7 · 6 · 5 · 4 · 3 choices.
If the number has 7 digits, then we have 7 · 7 · 6 · 5 · 4 · 3 · 2 choices.
If the number has 8 digits, then we have 7 · 7 · 6 · 5 · 4 · 3 · 2 · 1 choices.
(There cannot be a 9 digit number with distinct digits if there are no 2’s or 7’s.) Add up the numbers
above to get the grand total.
#7 Place a man (say the 1st man alphabetically). The rest of the table must go 2W, 1M, 2W, 1M, 2W,
1M, 2W. The number of ways to do this is:
8 · 7 · 3 · 6 · 5 · 2 · 4 · 3 · 1 · 2 · 1 = 8! · 3!
#10a: Number of ¡total
Number
of committees with one woman - Number of committees
¢ committees
¡12¢ ¡10¢ -¡10
¢
with no women = 22
−
·
−
=
23562.
OR
5
1
4
5
Number of 5 member committees with either 2, 3, 4, or 5 women:
µ ¶µ ¶ µ ¶µ ¶ µ ¶µ ¶ µ ¶µ ¶
12 10
12 10
12 10
12 10
+
+
+
.
2
3
3
2
4
1
5
0
#10b: Number of committees with man (not woman) + Number of committees with woman (not man) +
Number of committees with neither man nor woman =
·µ ¶µ ¶ µ ¶µ ¶ µ ¶µ ¶¸ ·µ ¶µ ¶ µ ¶µ ¶ µ ¶µ ¶ µ ¶µ ¶¸
11 9
11 9
11 9
11 9
11 9
11 9
11 9
+
+
+
+
+
+
+
2
2
3
1
4
0
1
3
2
2
3
1
4
0
·µ ¶µ ¶ µ ¶µ ¶ µ ¶µ ¶ µ ¶µ ¶¸
11 9
11 9
11 9
11 9
+
+
+
2
3
3
2
3
1
5
0
#14:
¡¢
Choose which seats the front row people occupy and arrange those people in those 5 seats: 85 · 5! = P (8, 5)
¡¢
Choose which seats the back row people occupy and arrange those people in those 4 seats: 84 · 4! = P (8, 4)
Choose¡ which
of the 7 unoccupied seats the other 5 people occupy and arrange those people in those 5
¢
7
seats: 5 · 5! = P (7, 5)
Multiply all these numbers together to get your answer: P (8, 5) · P (8, 4) · P (7, 5)
µ
#21a.
¶
9
9!
=
3, 2, 2, 1, 1
3!2!2!1!1!
#21b. You are either missing an S, E (same as missing a D), or an R (same as missing an A). The total
is therefore:
µ
¶
µ
¶
µ
¶
8
8
8
+2·
+2·
2, 2, 2, 1, 1
3, 2, 1, 1, 1
3, 2, 2, 1
#25a: Here is one way. There are many ways to solve this. I will give you my first method, which is rather
slow actually.
Choose a guy to sit in a car. Since all cars are the same, it does not matter which car he sits in.
Now choose the 3 people (of the 19 remaining) to sit with him and arrange all 4 people in his car, choose
the 4 people (of the µ
16 remaining)
and arrange them, .....
¶ in the next car
5
19
19!(4!)
1
This ends up being
(4!)5 =
= 19! · 4 = 20!. This final number makes sense because
4
3, 4, 4, 4, 4
3!(4!)
5
you can order all of the 20 people in a line (20! ways), put them in a car 4 at a time. Since it does not
matter which car is first, we can divide by 5.
¡ ¢
#25b: Choose 3 people to go in the car with Jerk 1 and arrange the people: 19
3 4!
Choose which
car
Jerk
2
is
in
and
choose
which
3
people
to
go
in
the
car
with
Jerk 1 and arrange the
¡ ¢
people: 4 · 15
4!
3
¡ 12 ¢
Arrange the other people as above: 4,4,4
(4!)3 )
¡19¢
¡15¢ ¡ 12 ¢
Grand total: 3 · 4 · 3 · 4,4,4 (4!)5
µ ¶
17
#28 : He has to go 17 blocks. 9 times he chooses to go east, 8 times north. Thus there are
ways to
8
do it without the flood.
If there is a flood, then he cannot go up 3, over 4 and then go on the east
µ ¶bound
µ ¶ street to that corner,
7
9
which is 5 blocks north and 4 blocks east of the office. Thus we get rid of
·
ways. Thus the total
3
5
number of legal routes is:
µ ¶ µ ¶ µ ¶
17
7
9
−
·
8
3
5