Permutations & Combinations:
Selected Exercises
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 ).
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10
There are 6 different candidates for governor.
In how many different orders can the names of the
candidates be printed on a ballot?
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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).
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20 (a)
How many bit strings of length 10 have exactly 3 0s?
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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
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20 (b)
How many bit strings of length 10 have more 0s than 1s?
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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.
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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?
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20 (b) Solution 2
1.
2.
There is a 1-to-1 correspondence between
•
strings with more 0s than 1s
•
strings with more 1s than 0s
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
10
20 (c)
How many bit strings of length 10 have ≥ 7 1s?
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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
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20 (d)
How many bit strings of length 10 have ≥ 3 1s?
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20 (d) Solution
Decompose this problem into disjoint sub-problems, & count each subproblem.
It is easier to:
1.
2.
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
subtract from the # of 10-bit strings (210):
The answer thus is 210 – (1 + 10 + 45) = 1024 – 56 = 968.
Copyright © Peter Cappello
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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.)
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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?
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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.
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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.
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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?
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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)
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20
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?
1
6
6
5
2
equivalent
A
5
B
4
3
4
1
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3
21
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:
1. Fix person1 at the head of the table: 1
2. Arrange the other 5 people at the table: 5!
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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)
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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?
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