# Basic counting principles_ day 1 by yurtgc548

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```									  Basic counting principles,
day 1
To decide how many ways something can
occur, you can draw a tree diagram.

Note that this only
shows half of the
tree – the automatic
transmission
n   Possible choices =      2      * 3 * 4 = 24
choices              transmission * music * color

n   Basic counting principle: If an event can
occur in p ways, and another event in q
ways, then there are p * q ways both
events can occur.
n   From now on we will use multiplication
and “fill the slots” as follows.
How many different batting orders are there in a 9-
person softball team? Fill the slots for each
position

1st    2nd    3rd    4th    5th    6th    7th    8th    9th
batter batter batter batter batter batter batter batter batter
9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1= 9!
1st      2nd      3rd      4th      5        6th      7th      8th      9th
batter   batter   batter   batter   batter   batter   batter   batter   batter

9 ways   8 ways   7 ways   Etc.
to       left,    now to
choose   once     chose
first    1st      3rd
batter   batter
chosen   batter

9! = 362,880 ways to write the batting order.
How many 7 digit phone numbers are there if
the first digit cannot be 0 or 1?

____*____*___* ____*____*____*____
8        *     10        *     10       * 10 * 10 * 10 * 10
# choices       # choices for all other digits will be the same
#                for
choices          second
for first          digit
digit

8,000,000 ways
How many 7-digit phone numbers begin with
867?

____*____ *____*____*____*____*____
1       * 1 * 1            * 10 * 10 * 10 * 10
one       one      one            these four digits can
choice    choice   choice           be 0-9 (10 choices)

10,000 different phone numbers
Using letters from the word “MATRIX”

How many 4-letter patterns can be formed?
# ways to fill slots:

_____ * _____ * _____ * _____
# ways to fill slots:

6 * 5 * 4 * 3 = 360
6 letters
to          5
choose      letters     etc.
from       left . .
Still using letters from the word
“MATRIX”, what if the first letter
must be a vowel?

_____ * _____ * _____ * _____
# ways to fill slots:

2 * 5 * 4 * 3 = 120
2
vowels       5
to    letters     etc.
choose    left . .
from
What if we fill only 4 slots and
the first and last letters must
be consonants?

______ * ______ * ______ * ______
# ways to fill slots:         4 * 4 * 3 * 3 = 144
4 conso-                           3 conso
nents to                            -nents
choose                              left to
fill these two first            from                             choose
from
4           3
letters     letters
left . .    left . .

fill less important
slots last
How many 3-digit palindromes
are possible? Note that a number
does not usually begin with a zero . . .

______ * ______ * ______
9 * 10 * 1 = 90
Can’t
begin                                   fill first
with a           must
zero             be the
and             same
be 3-            as the
digit           first
digit
(one
way to                       fill second
can     fill)
be
any
digit

then fill the last one
In a 5-card poker hand, the 1st 3 cards were
red face cards, the last 2 were black non-
face cards. How many ways can this
happen?

5 slots to fill:

_____ * ____ * _____ * _____* _____
6 * 5 * 4 * 20 * 19 = 45,600 ways
first 3 cards
choices:       A – 10 and
K, Q, J hearts       black: 20
or diamonds;        cards; 1st
1st slot, 2nd    slot, 2nd slot
slot, 3rd slot   (of that type)
Mutually Exclusive events
n   To get to school, Rita can either walk or ride the
bus. If she walks, she can take 3 routes, if she
rides the bus, there are 2 routes. Rita’s choices
are mutually exclusive - she can’t do both,
therefore, the possibilities are added to each
other.

n   3 ways to walk
n   2 ways to take a bus
n   total possible routes: 3 + 2 = 5
How many odd numbers between 10 and 1000
start and end with the same digit?

We have two mutually exclusive events:

___ * ___ two digit choices
+
___ * ___ * ___ three digit choices
two digit choices: just count multiples of 11
that are odd: (11, 33, 55, 77, 99) 5 choices

three digit choices: calculate like our
palindrome problem (3 slots to fill)
1 * 10 * 5 50 choices

total = 50 + 5 = 55 odd numbers between
10 and 1000 that have the same first and
last digit.
An example that is not mutually exclusive…
(each time you have the same number of
choices)

An ID label has 4 letters. How many
different labels are possible?

____ * _____ * _____ * _____
n   From before, we know it is
26* 26 * 26 *26     = 264 = 456,976

n   You can also look at it like:

4
26     where

26 (base) = number of different choices
4 (exponent) = number of times you make
that choice
On a multiple choice test with 15
questions and 4 answer choices per
combinations are there?

15
4 =1,073,741,824!!!!

4 = number of different choices
15 = number of times you make that choice
n   How many license plates of 2 symbols
(letters and digits) can be made using at
least one letter in each?
n   cases:
n   one letter:
L D or D L          2(26*10)

n   two letters:
n   LL           262
n   total: 2(26*10) + 262 = 1196 license
plates
How many ways can you make a 3 symbol
license plate using at least one letter?

What are the cases?
n   Cases if the license plate has at least one
letter:

n   one letter:
n   LDD or DLD or DDL            3(26*102)
n   two letters:                  +
n   L L D or L D L or D L L      3(262*10)
n   three letters:                    +
n   LLL                          263
Permutations

n   A permutation is an ordered arrangement:
ORDER MATTERS (KAT is different than
TAK). Our batting order problem and use
of the letters of matrix have been
permutations.
How many ways can letters in SPRING be
arranged?

6 * 5 * 4 * 3 * 2 * 1 = 6! =
720 ways to arrange the 6 letters when
order matters.

In general, there are n! permutations of n
objects.
What if we want to use the letters in
SPRING, but only want to know how many
2 letter arrangements can be made?
Using the “filling the slots” idea, we have

6 * 5 = 30 ways to fill two slots with
6 letter choices.
Another way to look at it is similar to
what we did for Pascal’s triangle:

In general, the number of permutations of
n objects taken r at a time is:
From 1000 contest entries, how many ways
can 1st, 2nd, and 3rd place prizes be
awarded? Does order matter?
We can fill in the slots:

_____ * _____ * _____

Or use our formula:
How does it change if the values
are not all unique?
How many 6-letter permutations of ACOSTA
are there?
note: there are two A’s that will not be
distinguishable from each other in a word.
A1A2COST = A2A1COST
the number of permutations will be cut in
half.
Is the bottom just “2” because there
are 2 A’s? What if there were more?
Consider using the letters of
PARABOLA:
Consider PARABOLA
the number of arrangements of the 3 A’s that will be
equivalent are:

A1A2A3PRBOL
A1A3A2PRBOL
A2A1A3PRBOL
A2A3A1PRBOL
A3A1A2PRBOL
A3A2A1PRBOL

This shows that 3! arrangements would be identical (the
number of ways you can arrange 3 objects in different
order). Thus, we need to divide by 3!, not just 3.
This generalizes to objects that have
more than one duplicate. If all objects
are used, the number of permutations
of n objects of which p and q represent
the number of items that are alike is:
How many permutations of the letters
in the word POSSIBILITY using only
5 letters are there?
where the denominator represents
the 2 S’s and the 3 I’s.
That’s all folks

Have a counting, permutable kind of day
J

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