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Unformatted text preview: ers without any restriction. Solution to Problem 7 Solution.
1 Since the rings are identical,
1 2 3 2 Number of distinct ways to wear the three rings on the same ﬁnger =
number of distinct ﬁngers = 8.
First we choose a ﬁnger to wear 2 rings, then we choose another ﬁnger
to wear 1 ring. Thus the number of ways = 8 × 7 = 56.
The number of ways to wear the three rings on three separate ﬁngers is
8
= 56.
3 Since the rings are distinguishable, the total number of distinct ways is
8 × 3! + 56 × 3! + 56 × 3! = 720. Problem 8 Problem 8. Mary has 5 dresses, 3 skirts, 4 blouses, 3 pairs of shoes,
and 2 hats. She always wears shoes and either a dress or a blouse and
a skirt. She may or may not wear a hat.
1
2 3 How many diﬀerent combinations can she wear?
Suppose Mary can aﬀord to buy either an extra dress or an extra hat
(but not both). Which should she buy if she decides to maximize the
number of diﬀerent combinations that she can wear?
Suppose Mary’s brown shoes do not match her pink or blue dress, and
that the blue hat does not match her yellow blouse. How many
matching combinations can she wear? Solution to Problem 8
Solution.
1
2 3 Total number of combinations = 3 × (2 + 1) × (5 + 3 × 4) = 153.
Extra dress: total number of combinations
= 3 × (2 + 1) × (6 + 3 × 4) = 162.
Extra hat: total number of combinations
= 3 × (3 + 1) × (5 + 3 × 4) = 204.
Therefore she should buy an extra hat.
Total number of combinations with brown shoes and pink or blue dress
= 1 × 3 × 2 = 6.
Total number of combinations with blue hat and yellow blouse
= 3 × 1 × (1 × 3) = 9.
Since the two events “brown shoes and pink or blue dress” and “blue
hat and yellow blouse” are mutually exclusive, there is no double count
in the above two countings. Therefore the matching combinations she
can wear is
153 − 6 − 9 = 138. Problem 9 Problem 9 (a more diﬃcult question). Suppose there are N boys and
M girls sitting around a table, and the chairs are all diﬀerent.
1
2 How many diﬀerent ways to sit them?
How many ways to sit them such that there are k boys with a girl
sitting to his right (assume M ≥ k )? Solution to Problem 9 I
Solution
1
2 (M + N )!.
We can count the total number of combinations using the following
steps:
1 2 3
4 Suppose there is a boy called John. Line up the boys with John in
front. The number of ways to line up the N − 1 boys is (N − 1)!.
Choose k boys from the N boys and mark them. The number of ways
N
is
.
k
Line up the M girls. The number of ways is M !.
Divide the M girls (which are already all lined up) into k groups
M −1
without disturbing the order. The number of ways is
(why?
k −1
Consider for example there are 4 girls lined up in a row G1 G2 G3 G4 , how
many ways to separate them into 3 groups without disturbing the
order?). Put the ﬁrst group on the right of the ﬁrst marked boy, and
put the 2nd group on the right of the 2nd marked boy, etc. Solution to Problem 9 II 5 Choose a sit for John. People queuing behind John occupy seats to
John’s right one by one anticlockwise. There are M + N ways.
Total number of ways
= the product of the number of ways in (i), (ii), (iii), (iv), (v)
=(N − 1)!
=N !M ! N
k
N −1
k −1 M! M −1
k −1
M −1
k −1 (M + N )
M +N
k...
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This note was uploaded on 01/16/2012 for the course STAT 1301 taught by Professor Smslee during the Fall '08 term at HKU.
 Fall '08
 SMSLee
 Statistics, Probability

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