Last time we discussed:

1

• Last time we discussed
• The rule of sum
• The rule of product
• Permutations arrangements when the order is
  important
• all objects are distinct/not all are distinct
• without repetition/ with repetitions

• Combinations selections when the order does not
  matter

2
Suppose two dice are rolled, one blue and one red.

• How many outcomes are possible?

Product rule 6?636

• How many outcomes are doubles?

The rule of sum 6

• How many outcomes give the sum 7 or the sum 11?

16 25 34 43 52 61
56 65
3

• How many outcomes have the blue die showing 2?

6

• How many outcomes have exactly one die showing 2?

55

• How many outcomes have at least one die showing 2?

101

• How many outcomes have neither die showing 2?

5?5
36?11
4

• How many outcomes give an even sum?

Possible even sums 12, 10, 8, 6, 4, 2.
1055 46 64
862 53 44 35 26
1266
651 42 33 24 15
431 22 13
211
5
Suppose you have 12 students in the recitation
class who are supposed to break up into 4 groups
of 3 students each. How many different group
assignments are possible? (There is no
numbering of the groups. All that matters is
who collaborates with whom).
There are 12! different lists (permutations), but
not all of them result in a different group
assignment.
How many different permutations represent the
same group assignment?
We can also permute different groups in 4! ways.
6
So, there are 12! different lists, and each
assignment is produced by (3!)4?4! different
lists.
Ans 12!/((3!)4?4!)
7
And you need to take into account that groups are
not distinguishable.
8
Combinations with repetitions.
Consider 4-combinations of a,b a, a, a, a,
a, a, a, b, a, a, b, b, a, b, b, b, b, b,
b, b
9
Combinations with repetitions.
Donut shop has 5 types of donuts. In how many
ways we can select ten donuts?
This problem can be represented as an equivalent
arrangement of ten donuts into 5 boxes. All
possible distributions Can be considered as
permutations of a dozen of donuts and 4
separators between boxes
10
We need to count the number of permutations of 10
donuts and 4 separators. So, we have 14 objects,
4 of which are identical and 10 are identical.
11
The number of r-combinations of n objects that
can be repeated (any number of times)
can be considered as the number of arrangements
of r identical objects and n-1 separators
(bars).
12
How many distinct integer solutions of the
equation xyz10 exist if x, y, z?0?
Any solution corresponds to some distribution of
10 between three distinct boxes
How to count all possible arrangements of 10
stars and 2 bars?
13
Pigeonhole Principle.
If k 1 or more pigeons fly into k pigeonholes,
then there is at least one pigeonhole containing
two or more pigeons.
Examples. Among any group of 366 people there
must be at least two with the same birthday,
because there are only 365 possible birthdays.
In any group of 27 words there must be at least
two that begin with the same letter.
How many students must be in class to guarantee
that at least two students receive the same
score on the final (exam is graded on a scale
from 0 to 100)? Ans. 102
14
Generalized Pigeonhole principle.
If n objects are placed into k boxes, then there
is at least one box containing at least ?n/k?
objects. Here ?n/k? denotes the largest
integer less then n/k1.
Among 100 people there are at least ?100/12?
?8.333? 9 who were born in the same month.
How many students, each of whom comes from one of
50 states, should be enrolled in a university to
ensure that there are at least 10 coming from
the same state?
451