Generalized Permutations & Combinations:
Selected Exercises
10 (a)
A croissant shop has plain croissants, cherry
croissants, chocolate croissants, almond
croissants, apple croissants, & broccoli croissants.
How many ways are there to choose 12 croissants?
Abstract version of the problem
There is an infinite supply of 6 kinds of objects.
How many ways are there to choose 12 of them?
2
10 (a) Solution
How many binary strings of length 12 + 6 – 1 are there with
exactly 12 0s? (= # of length-17 binary strings with 5 1s.)
C( 12 + 6 – 1, 6 – 1 ) = C( 12 + 6 – 1, 12 ).
How many ways are there to order 12 items from a menu of 6
kinds of items?
1-to1 correspondence between binary strings and orders.
Item 1 Item 2 Item 3 Item 4 Item 5 Item 6
3
10 (b)
A croissant shop has plain croissants, cherry croissants,
chocolate croissants, almond croissants, apple croissants, &
broccoli croissants.
How many ways are there to choose 36 croissants?
Abstract version of the problem
There is an infinite supply of 6 kinds of objects.
How many ways are there to choose 36 of them?
4
10 (b) Solution
How many binary strings of length 36 + 6 – 1 are
there with exactly 36 0s? (= # of length-41 binary
strings with 5 1s.)
C( 36 + 6 – 1, 6 – 1 ) = C( 36 + 6 – 1, 36 ).
How many ways are there to order 36 items from a
menu of 6 kinds of items?
1-to1 correspondence between binary strings and orders.
5
10 (c)
A croissant shop has plain croissants, cherry croissants, chocolate
croissants, almond croissants, apple croissants, & broccoli croissants.
How many ways are there to choose 24 croissants with ≥ 2 of each kind?
Abstract version of the problem
There is an infinite supply of 6 kinds of objects.
How many ways are there to choose 24 of them with ≥ 2 of each kind?
6
10 (c) Solution
How many ways are there to order 24 – 2 . 6 = 12
items from a menu of 6 kinds of items?
There is a 1-to-1 correspondence between these
orders and the original type of orders:
For each order of 12 items:
For each kind of item, increment the order of that kind by 2.
The resulting order has 24 items, ≥ 2 of each kind of item.
Answer: C( 12 + 6 - 1, 6 – 1 ).
7
10 (d)
A croissant shop has plain croissants, cherry croissants, chocolate croissants,
almond croissants, apple croissants, & broccoli croissants.
How many ways are there to choose 24 croissants with  2 broccoli?
Abstract version of the problem
There is an infinite supply of 6 kinds of objects.
How many ways are there to choose 24 of them with  2 of kind 1?
8
10 (d) Solution
We can use the sum rule to decompose this problem
into 3 sub-problems, based on an exact # of
broccoli croissants:
1. Count the solutions with exactly 0 broccoli croissant
C( 24 + 5 - 1, 5 – 1 ).
2. Count the solutions with exactly 1 broccoli croissant
1. Pick the broccoli croissant: 1
2. Pick the 23 remaining croissants from the remaining 5 kinds of
croissants: C( 23 + 5 - 1, 5 – 1 ).
3. Count the solutions with exactly 2 broccoli croissants
1. Pick the 2 broccoli croissants: 2
2. Pick the 22 remaining croissants from the remaining 5 kinds of
croissants: C( 22 + 5 - 1, 5 – 1 ).
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10 (d) Better Solution
1. Count all orders of 24 croissants:
C( 24 + 6 – 1, 6 – 1 )
2. Subtract the “bad” orders:
Order 21 other croissants from all 6 varieties:
C( 21 + 6 – 1, 6 – 1 )
Answer: C( 24 + 6 – 1, 6 – 1 ) – C( 21 + 6 – 1, 6 – 1 )
10
10 (e)
A croissant shop has plain croissants, cherry croissants, chocolate croissants,
almond croissants, apple croissants, & broccoli croissants.
How many ways are there to choose 24 with ≥ 5 chocolate & ≥ 3 almond?
Abstract version of the problem
There is an infinite supply of 6 kinds of objects.
How many ways are there to choose 24 of them with ≥ 5 of kind 1 & ≥ 3 of kind 2?
11
10 (e)
There is a 1-to-1 correspondence between
1. orders of 24 – 5 – 3 objects of 6 kinds
2. Orders of 24 objects of 6 kinds with ≥ 5 of kind 1 & ≥ 3 of kind 2.
Correspondence: Add
5 objects of kind 1
3 objects of kind 2
to each order of type 1 to get an order of type 2.
We count the # of orders of type 1:
Answer: C( 24 – 5 – 3 + 6 – 1, 6 – 1 )
12
10 (f)
A croissant shop has plain croissants, cherry croissants, chocolate croissants, almond croissants, apple
croissants, & broccoli croissants.
How many ways are there to choose 24 with:
•
≥ 1 plain & ≥ 2 cherry & ≥ 3 chocolate & ≥ 1 almond & ≥ 2 apple
•
 3 broccoli?
Abstract version of the problem
There is an infinite supply of 6 kinds of objects.
How many ways are there to choose 24 of them with:
–
≥ 1 of kind 1 & ≥ 2 of kind 2 & ≥ 3 of kind 3 & ≥ 1 of kind 4 & ≥ 2 of kind 5
–
 3 of kind 6
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10 (f)
1. Put 1 plain & 2 cherry & 3 chocolate & 1 almond
& 2 apple to the side (9 objects to the side)
2. There are 24 – 9 = 15 left to distribute w/o
restriction on broccoli.
C( 15 + 6 – 1, 6 – 1 )
3. Subtract the bad orders (≥ 4 broccoli):
C( 11 + 6 – 1, 6 – 1)
Answer: C( 15 + 6 – 1, 6 – 1 ) – C( 11 + 6 – 1, 6 – 1)
14
20
How many integer solutions are there to the inequality
x1 + x2 + x3  11, for x1 , x2 , x3 ≥ 0?
Hint: Introduce an auxiliary variable x4 such that
x1 + x2 + x3 + x4 = 11.
15
20 Solution
There is a 1-to-1 correspondence between:
solutions to the equality
solutions to the inequality question
The values of x1 , x2 , & x3 from the equality solve the
inequality.
Equivalent to counting solutions to the equality:
How many ways are there to order 11 items from a menu of
4 kinds of items?
Answer: C( 11 + 4 – 1, 4 – 1 ).
16
30
How many different strings can be made from the
letters in MISSISSIPPI, using all 11 letters?
17
30 Solution
Using the product rule:
1. Pick the position in the 11-letter string where the
letter “M” goes: C( 11, 1 )
2. Pick the position in the 10 remaining positions
where the 4 “I” letters go: C( 10, 4 )
3. Pick the position in the 6 remaining positions
where the 4 “S” letters go: C( 6, 4 )
4. Pick the position in the 2 remaining positions
where the 2 “P” letters go: C( 2, 2 )
Answer: 11! / 1!4!4!2!.
18
30 Generally
If you have n objects such that:
n1 objects of them are of type t1
n2 objects of them are of type t2
...
nk objects of them are of type tk
The # of arrangements of these objects is
C(n, n1 ) C( n - n1 , n2 ) C( n – n1 – n2 , n3 ) … C( n – n1 – … – nk-1,
nk )
= n! / (n1! n2! … nk!)
(This equality is simple to verify algebraically.)
19
40
How many ways are there to travel in xyzw space
from the origin (0, 0, 0, 0) to (4, 3, 5, 4) by taking
steps:
1 unit in the positive x direction
1 unit in the positive y direction,
1 unit in the positive z direction
1 unit in the positive w direction?
20
40 Solution
Any path from (0, 0, 0, 0) to (4, 3, 5, 4) is a sequence with:
4 x steps
3 y steps
5 z steps
4 w steps.
Equivalent problem: How many 16 letter sequences of x, y, z,
and w are there with exactly
4 x,
3y
5z
4w?
Answer: 16! / 4! 3! 5! 4!.
21
50
How many ways are there to distribute 5
distinguishable objects in 3 indistinguishable
boxes?
22
50 Solution
Use the sum rule to decompose the set of solutions into
disjoint subsets:
# solutions with 5 objects in 1 box: 1.
# solutions with 4 objects in 1 box, 1 object in another: C(5, 1).
# solutions with 3 objects in 1 box, 2 objects in a 2nd: C(5, 3).
# solutions with 3 objects in 1 box, 1 object in a 2nd box, 1
object in a 3rd box: C(5, 3).
# solutions with 2 objects in 1 box, 2 objects in a 2nd box, 1
object in a 3rd box: C(5, 1) C(4, 2).
Answer: 1 + C(5, 1) + C(5, 3) + C(5, 3) + C(5, 1) C(4, 2).
23
60
Suppose a basketball league has 32 teams, split into 2
conferences of 16 teams each. Each conference is split into
3 divisions. Suppose that the North Central Division has 5
teams. Each of the teams in this division plays:
4 games against each of the other teams in this division
3 games against each of the 11 remaining teams in the division, and
2 games against each of the 16 teams in the other conference.
In how many different orders can the games of 1 of the teams
in the North Central Division be scheduled?
24
60 Solution
Let the 4 other teams in the North Central Division be named
x1, x2, x3, x4.
Let the 11 other teams in the division be named y1, y2, …, y11.
Let the 16 teams in the other division be named z1, z2, …, z16.
The total # of games that a team plays is
4 . 4 + 11 . 3 + 16 . 2 = 81
The number of 81-letter sequences with:
4 each of x1, x2, x3, x4
3 each of named y1, y2, …, y11
2 each of z1, z2, …, z16 is
81! / (4!)4(3!)11(2!)16
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