In how many ways can you choose an odd number from 1 to 12?

  Answer:  How many odd numbers are there from 1 to 12?  6
                 So there are 6 ways to choose an odd number.

In how many ways can the sum 13 be obtained?

  Answer:  A sum of 13 can be obtained in the following ways:
                  1 + 12, 2 + 11, 3 + 10, 4 + 9, 5 + 8, 6 + 7
                  So there are 6 ways to obtain a sum of 13.
 

Fundamental Counting Principle:

If one event can occur in M ways and a second event in N ways, then the number of ways the two events can occur is MN ways.

In how many ways can a 6-question true-false test be answered?
  Answer:  There are 2 ways to answer each question.  There are 6 questions.  So there are
                 2(2)(2)(2)(2)(2) or 26 ways to answer the questions.
 

Permutations:

A permutation is an ordering of elements.  Order is important.  There are n! ways to order n objects.  The number of permutations of n objects taking r at a time is nPr  =       n!
                                                                                   (n - r)!

How many 3-digit numbers can be formed under the following conditions?
  (1)  The leading digit cannot be zero.
         Answer:  _ _ _ represents the 3-digit number.  The number of digits that can be used in each blank is as
                        follows:  9  10  10 .  So there are 900 ways to do this.

  (2)  The leading digit cannot be zero and no repetition of digits is allowed.
         Answer: 9  9  8   So there are 648 ways to do this.

  (3)  The leading digit cannot be zero and the number must be a multiple of 5.
         Answer: 9  9  2   So there are 162 ways to do this.  (Note:  a multiple of 5 must end in 0 or 5.)

  (4)  The number is at least 400.
         Answer:  6 9  9    So there are 486 ways to do this.  (Note:  the first number must be 4,5,6,7,8, or 9)

Find the number of distinguishable permutations of the letters in Eli Harrington.
 

Combinations:

     nCr =       n!
               (n - r)! r!

An employer interviews 8 people for 4 openings in the company.  3 of the 8 people are women.  If all 8 are qualified, in how many ways can the employer fill the 4 positions if

(a) the selection is random   (Note:  4 people are being chosen from 8)

          8C4 =       8!         =          8!      =      8(7)(6)(5)(4)(3)(2)(1)    =      8(7)(6)(5)   =     (7)(2)(5)  =  70
                   (8 - 4)! 4!              4! 4!           4(3)(2)(1)(4)(3)(2)(1)            4(3)(2)(1)
 

and (b) exactly two women are selected?
            (Note:  2 out of 3 women are chosen, while 2 out of 5 men are chosen)

           (3C2 ) ( 5C2 ) =      3!                  5!   =             3!5!        =      3(2)(1)(5)(4)(3)(2)(1)    =      3(5)(2)   =    30
                                  (3 - 2)! 2!    (5 - 2)!2!           1!2!3!2!           1(2)(1)(3)(2)(1)(2)(1)


Problems(Questions are there, but no answers are posted)


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