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Basic Mathematics-II




                    Notes          If you don’t contain a factorial key, you can abridge it as exposed above and then enter it in.  It
                                   is most likely best to shorten it first, because in some cases the numbers can get somewhat large,
                                   and it would be awkward to multiply all  those numbers one by one.
                                   This shows there are 30 draws that would contain 1 RED and 2 WHITE marbles.
                                   How many different draws would contain exactly 2 red marbles?
                                   This  would be a combination problem, since a draw would be a group of  marbles without
                                   considering  order.  It is like taking hold of a handful of marbles and gazing at them.

                                   In part a above, we gazed  at all possible draws.  From that  list we  only desire the ones that
                                   enclose 2 RED and 1 WHITE marbles.  Memorize that we require a total of 3 marbles in the
                                   draw.  As we have to have 2 red, that makes us needing 1 white to complete the draw of 3.
                                   Let’s observe what the draw appears like:  we would have to have 2 red and 1 white marbles to
                                   fulfill this condition:
                                   2 RED    1 WHITE

                                   First we need to find n and r:
                                   Mutually that would make up 1 draw.  We will use the counting principle to assist us with this one.
                                   Note how 1 draw is split into two parts - red and white.  We can not merge them together since
                                   we require a specific number of each one. So we will work out how many ways to get 2 RED and
                                   how many ways to get 1 WHITE, and by means of the counting principle, we will multiply these
                                   numbers together.
                                   2 RED:
                                   If n is the number of RED marbles we have to select from,  what do you think n is in this problem?

                                   If you said n = 3 you are correct!!!  There are a total of  3 RED marbles.
                                   If r is the number of RED marbles we are drawing at a time, what do you think r is?
                                   If you said r = 2, pat yourself on the back!! 2 RED marble is drawn at a time.
                                   1 WHITE:
                                   If n is the number of WHITE marbles we have to select from,  what do you think n is in this problem?
                                   If you said n = 5 you are right!!!  There are a total of 5 WHITE marbles.
                                   If r is the number of WHITE marbles we are drawing at a time, what do you think r is?

                                   If you said r = 1, tap yourself on the back!! 1 WHITE marble are drawn at a time.
                                   Let us place those values into the combination formula and see what we obtain:


                                                                                   r
                                                                              n
                                                          ! n            *RED:   = 3,   = 2
                                                   C 
                                                  n  r
                                                                                     r
                                                                                 n
                                                         
                                                       (n r )! !         *WHITE:   = 5,   = 1
                                                            r
                                                            3!     5!
                                                   C  . C      .
                                                  3  2 5  1
                                                             )
                                                              !
                                                                   
                                                                     !
                                                           
                                                         ( 3 2 2! ( 5 1 1!
                                                                     )
                                                    3!  5                *Eval. inside ( )
                                                     .
                                                        !
                                                    2 1 4 1!
                                                      !
                                                     !
                                                                         *Expand 3! until it gets to 2!
                                                    3 2 5 4!
                                                      !
                                                        .
                                                    .
                                                                          *Expand 5! until it gets to 4!
                                                    2 1 4 1!
                                                     !
                                                        !
                                                      !
                                                    3 5
                                                    .
                                                    1 2
                                                   15
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