5
For type I, there are C(2,1) x C(3,1) x
5
4
!
!
such “words”.
For type II, there are C(2,1) x C(2,1) x
5
3 2
!
! !
such “words”.
For type III, there are C(2,1) x C(3,2) x
5
3
!
!
such “words”.
For type IV, there are C(3,2) x C(2,1) x
5
2 2
!
! !
such “words”.
For type V, there are C(3,1) x
5
2
!
!
such “words”. Using the “or” principle,
our answer will be the sum of all these cases.
8.
page 189, 12
A standard 52 card deck has 13 types of cards, each type comes in four
suits.
a.
The number of five-card poker hands with four aces, is C(4,4) x
C(48,1). C(4,4) counts the number of ways to select four aces out of the
four aces available. After the four aces are selected there are C(48,1)
ways to select a non-ace.
b.
The number of five-card poker hands with four of a kind is
C(13,1)xC(4,4)xC(48,1). From thirteen types of cards, pick one of
them, i.e. C(13,1) ways, next pick four of the four cards of available of
that type, i.e. C(4,4) ways, and with the forty-eight cards left pick one
them, i.e. C(48,1) ways.
c.
The number of five-card poker hands with exactly two different pairs is
C(13,2) x C(4,2) x C(4,2) x C(44,1). From thirteen types of cards, pick
two of them, i.e. C(13,2) ways, next pick two of the four cards available
of the first type, do the same with the second type, i.e. C(4,2) x C(4,2)
ways to do both, and with the forty-four cards remaining select one
more ,i.e. C(44,1) ways.