AMS 301.2
(Spring 2011)
Homework Set # 7 – Solution Notes
# 14, 5.1:
Here we must select in which of the 6 places we should place the “2”, and then in
which of the 5 remaining places we should place the “3”. The 4 “1”s will be placed in the remaining
4 spots. So the answer is 6
·
5 = 30.
This is equivalent to looking at how many different “spellings” there are using the characters
“111123”:
6!
4!1!1!
# 18, 5.1:
(a). The letters can appear first or last, 2 choices. The number of ways of chooosing
3 letters, repetition allowed and order is important is 26
3
, similarly 3 numbers 10
3
, so the answer
is 2
·
26
3
·
10
3
.
(b). The number of choices for the letters are (26 + 26
2
+ 26
3
). If there is a single digit, the
letters can appear either before it or after, so there are 2
·
10 choices. If there are 2 digits, then
the placement of the letters has 3 choices, and there are 100 possibilities for the digits. The final
answer is: (26 + 26
2
+ 26
3
)(2
·
10 + 3
·
10
2
+ 4
·
10
3
).
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '11
 Estie
 Playing card, ﬁrst player, ﬁrst spade card

Click to edit the document details