Permutations & Combinations: Selected Exercises. Copyright © Peter Cappello2 Preliminaries Denote the # of arrangements of some k elements of a set of

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 ).


10

There are 6 different candidates for governor.

In how many different orders can the names of the candidates be printed on a ballot?


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)





The composite number is 6! = 6 . 5 . 4 . 3 . 2 . 1 = 720.

This also is known as P(6,6).


20 (a)

How many bit strings of length 10 have exactly 3 0s?


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.


20 (b)

How many bit strings of length 10 have more 0s than 1s?


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.


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?


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.


20 (c)

How many bit strings of length 10 have ≥ 7 1s?


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.


20 (d)

How many bit strings of length 10 have ≥ 3 1s?


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.


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.)


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?


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



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.


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.


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?


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)


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


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!


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)


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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Permutations & Combinations: Selected Exercises. Copyright © Peter Cappello2 Preliminaries Denote the # of arrangements of some k elements of a set of