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
 question, how many answer
 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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