CSE21 FA11
Homework #2
(10/3/11)
In this homework, we will consider “ordinary” decks of playing cards which
have 52 cards, with 13 of each of the four suits (Hearts, Spades, Diamonds
and Clubs) with each suit have the 13 ranks (Ace, 2, 3, . . . , Jack, Queen,
King). (See the text if you need more details).
2.1.
(a) How many different rearrangements of all the letters in
UNSCRUPULOUS
are there?
(b) How many different rearrangements of all the letters in
supercalifragilisticexpialidocious
are there?
Solution:
(a) Since there are 1 ‘C’, 1 ‘L’, 1 ‘N’, 1 ‘O’, 1 ‘P’, 1 ‘R’, 2 ‘S’, and 4 ‘U’, the
number of different rearrangements is
12
1
,
1
,
1
,
1
,
1
,
1
,
2
,
4
.
(b) Since there are 3 ‘a’, 3 ‘c’, 1 ‘d’, 2 ‘e’, 1 ‘f’, 1 ‘g’, 7 ‘i’, 3 ‘l’, 2 ‘o’, 2 ‘p’, 2
‘r’, 3 ‘s’, 1 ‘t’, 2 ‘u’ and 1 ‘x’, the number of different rearrangements is
34
3
,
3
,
1
,
2
,
1
,
1
,
7
,
3
,
2
,
2
,
2
,
3
,
1
,
2
,
1
.
2.2.
A 20member club must have a president, vicepresident, secretary,
and a treasurer, as well as a threeperson nominating committee.
If the
officers must be different people, and if no officer may be on the nominating
committee, in how many ways could the officers and nominating committee
be chosen? Answer the same question if officers may be on the nominating
committee.
Solution:
1
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(a) Note that there are 20
4
ways to choose 4 members out of 20 members
as president, vicepresident, secretary, and treasurer respectively. Then, for
the left 16 members, there are
(
16
3
)
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 Fall '07
 Graham
 Treasurer, Playing card, vicepresident, 5card hands

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