Permutations & Combinations: Selected Exercises. Copyright © Peter Cappello2 Preliminaries Denote...
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Transcript of Permutations & Combinations: Selected Exercises. Copyright © Peter Cappello2 Preliminaries Denote...
Permutations & Combinations: Selected Exercises
Copyright © Peter Cappello 2
Preliminaries
Denote the # of arrangements of some k elements of a
set of n elements as P( n, k ).
Use the product rule to derive a formula for P( n, k ).
Let C( n, k ) be the # of subsets of k elements drawn
from a set of n elements.
Use the product rule to derive a formula for C( n, k )
in terms of P( n, k ) & P( k, k ).
Copyright © Peter Cappello 3
10
There are 6 different candidates for governor.
In how many different orders can the names of the candidates be printed on a ballot?
Copyright © Peter Cappello 4
10 Solution
The # of different orders that the candidate names can be printed on a
ballot is described by the following procedure:
1. Pick the candidate that appears on top (6)
2. Pick the candidate that appears below that (5)
3. Pick the candidate that appears below that (4)
4. Pick the candidate that appears below that (3)
5. Pick the candidate that appears below that (2)
6. Pick the candidate that appears below that (1)
The composite number is 6! = 6 . 5 . 4 . 3 . 2 . 1 = 720.
This also is known as P(6,6).
Copyright © Peter Cappello 5
20 (a)
How many bit strings of length 10 have exactly 3 0s?
Copyright © Peter Cappello 6
20 (a) Solution
The bit strings have 10 positions: 1, 2, …, 10.
A bit string with exactly 3 0s can be described as a 3-subset of the
numbers 1, 2, …, 10.
These are the bit positions where the 0s go.
There are C(10, 3) such 3-subsets.
For each such 3-subset, all other positions take 1s.
There is 1 way to do that.
The answer thus is C(10, 3) = 10 . 9 . 8 / 3 . 2 . 1 = 120.
Copyright © Peter Cappello 7
20 (b)
How many bit strings of length 10 have more 0s than 1s?
Copyright © Peter Cappello 8
20 (b) Solution 1
Decompose this problem into disjoint sub-problems; count each
sub-problem:
1. 6 0s & 4 1s: C(10, 6) = C(10, 4) = 10.9.8.7 / 4.3.2 = 210
2. 7 0s & 3 1s: C(10, 7) = C(10, 3) = 10 . 9 . 8 / 3 . 2 = 120
3. 8 0s & 2 1s: C(10, 8) = C(10, 2) = 10 . 9 / 2 = 45
4. 9 0s & 1 1: C(10, 9) = C(10, 1) = 10
5. 10 0s & 0 1s : C(10, 10) = C(10, 0) = 1
The answer thus is
C(10, 4) + C(10, 3) + C(10, 2) + C(10, 1) + C(10, 0)
= 210 + 120 + 45 + 10 + 1 = 386.
Copyright © Peter Cappello 9
20 (b) Solution 1
Is the following analysis right?
1. Pick the positions of 6 0s:
C(10, 6) = C(10, 4)
2. Fill in the other 4 positions: 24
C(10, 4) 24 = 3,360 386.
What is wrong?
Copyright © Peter Cappello 10
20 (b) Solution 2
1. There is a 1-to-1 correspondence between
• strings with more 0s than 1s
• strings with more 1s than 0s
2. Strategy:
1. C(10, 5) = the # of strings with an equal # of 1s & 0s.
2. 210 – C(10, 5) = the # with an unequal # of 1s & 0s.
3. (210 – C(10, 5) ) / 2 = the # with more 0s than 1s.
C(10, 5) = 10.9.8.7.6 / 5.4.3.2.1 = 252
(1024 – 252)/2 = 386.
Copyright © Peter Cappello 11
20 (c)
How many bit strings of length 10 have ≥ 7 1s?
Copyright © Peter Cappello 12
20 (c) Solution
Decompose this problem into disjoint sub-problems, and count
each sub-problem:
1. 7 1s & 3 0s: C(10, 7) = C(10, 3) = 10 . 9 . 8 / 3 . 2 = 120
2. 8 1s & 2 0s : C(10, 8) = C(10, 2) = 10 . 9 / 2 = 45
3. 9 1s & 1 0: C(10, 9) = C(10, 1) = 10
4. 10 1s & 0 0s : C(10, 10) = C(10, 0) = 1
The answer thus is
C(10, 3) + C(10, 2) + C(10, 1) + C(10, 0) = 120 + 45 + 10 + 1 = 176.
Copyright © Peter Cappello 13
20 (d)
How many bit strings of length 10 have ≥ 3 1s?
Copyright © Peter Cappello 14
20 (d) Solution
Decompose this problem into disjoint sub-problems, & count each sub-
problem.
It is easier to:
1. count the number of 10-bit strings w/o the property
1. 0 1s & 10 0s: C(10, 0) = 1
2. 1 1 & 9 0s: C(10, 1) = 10
3. 2 1s & 8 0s: C(10, 2) = 45
2. subtract from the # of 10-bit strings (210):
The answer thus is 210 – (1 + 10 + 45) = 1024 – 56 = 968.
Copyright © Peter Cappello 15
30 (a)
There are 7 women & 9 men.
How many ways are there to select a committee of
5 members, with at least 1 woman?
(In such problems, it is customary & implicit to
take individuals as distinct.)
Copyright © Peter Cappello 16
30 (a)
Consider using the product rule:
1. Pick 1 woman: C(7,1).
2. Pick 4 members from the remaining 6 women & 9
men: C(15,4).
Is the answer: C(7,1) C(15,4)?
Given a committee of men & women, can you
identify the stage at which each woman was
chosen?
Copyright © Peter Cappello 17
30 (a) Solution
Decompose the problem into disjoint sub-problems:
1. The committee has 1 woman:
1. Pick the woman: C(7, 1) = 7
2. Pick the men: C(9, 4) = 9 . 8 . 7 . 6 / 4 . 3 . 2 = 126
2. The committee has 2 women:
1. Pick the women: C(7, 2) = 7 . 6 / 2 = 21
2. Pick the men: C(9, 3) = 9 . 8 . 7 / 3 . 2 = 84
3. The committee has 3 women: C(7, 3) . C(9, 2) = 35 . 36
4. The committee has 4 women: C(7, 4) . C(9, 1) = 35 . 9
5. The committee has 5 women: C(7, 5) . C(9, 0) = 21 . 1
The answer is
C(7, 1)C(9, 4) + C(7, 2)C(9, 3) + C(7, 3)C(9, 2) + C(7, 4)C(9, 1) + C(7,5)C(9, 0)
= 7 . 126 + 21 . 84 + 35 . 36 + 35 . 9 + 21 . 1 = 4,242.
Copyright © Peter Cappello 18
30 (a) More Elegant Solution
The set of all committees with 5 members is the universe. Its
size is C(7 + 9, 5).
Subtract all committees w/o women: C(9, 5).
The answer is C(16, 5) – C(9, 5) = 4,368 – 126 = 4,242.
Copyright © Peter Cappello 19
30 (b)
There are 7 women & 9 men.
How many ways are there to select a committee of 5
members, with ≥ 1 woman & ≥ 1 man?
Copyright © Peter Cappello 20
30 (b) Solution
Subtract “bad” committees from all 5-committees:
1. The # of all 5-committees: C(16, 5)
2. The # of 5-committees w/o women: C(9, 5)
3. The # of 5-committees w/o men: C(7, 5)
The answer: C(16, 5) – C(9, 5) – C(7, 5)
Copyright © Peter Cappello 21
40
How many ways are there to seat 6 people around a
circular table, where 2 seatings, A & B, are equivalent
if A is a rotation of B?
A
1
5
6 2
3
4
B
6
4
5 1
23
equivalent
Copyright © Peter Cappello 22
40 Solution
If the people sat in a line the answer is 6!
If we drag the line seating into a circle, 6 rotations
(permutations) of that “line seating” are equivalent.
The answer is 6!/6 = 5!
The equivalence relation has 5! equivalence classes, each with 6
elements.
Alternatively:
– Fix person1 at the head of the table: 1
– Arrange the other 5 people at the table: 5!
Copyright © Peter Cappello 23
Computing C(n,k)
How many ways are there to select a team of k players
from a set of n players, with a particular player named
as captain?
1. Pick the k players: C(n, k)
2. Pick the captain: C(k, 1) = k
Equivalently,
1. Pick the captain: C(n, 1) = n
2. Pick the remainder of the team: C(n-1, k-1)
Copyright © Peter Cappello 24
Computing C( n, k )
C( n, k )k = n C( n-1,k-1 )
C( n, k ) = n/k C( n-1, k-1 ).
Apply the above recursively:
C(n,1) = n, for k = 1
C( n, k ) = n(n-1) . . . (n - k +1) /k! , for k > 1.
For example,C(1000, 4) = 1000 . 999 . 998 . 997 / 4 . 3 . 2 . 1
Why does each factor in the denominator divide some factor in the numerator?