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Basic Mathematics-II




                    Notes              If you contain a factorial key, you can place it in as 8!, divided by 5!, divided by 3! and then
                                       press enter or =.
                                       If you don’t have a factorial key, you can make it simpler as revealed above and then enter
                                       it in.  It is most likely best to shorten it first, since in some cases the numbers can get rather
                                       large, and it would be awkward to multiply all those numbers one by one.

                                       This indicates that there are 56 different draws.
                                   (b)  How many different draws would contain only red marbles?

                                       This would be a combination problem, since a draw would be a group of marbles without
                                       regard to order.  It is like clutching a handful of marbles and gazing at them.
                                       In part a above, we observed all possible draws.  From that list we only desire the ones
                                       that include only red.
                                       Let us observe  what the draw appears like:  we would have to have 3 red marbles to fulfill
                                       this condition:
                                       3 RED
                                       First we are required to find n and r :

                                       If n is the number of RED marbles we have to select from, what do you think n here?
                                       If you said n = 3 you are right!!!  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 = 3, tap yourself on the back!! 3 RED marbles are drawn at a time.
                                       Let us place those values into the combination formula and observe what we obtain:
                                                                    ! n
                                                             C          *  = 3,   = 3
                                                                          n
                                                                               r
                                                            n  r
                                                                  
                                                                      r
                                                                (n r )! !
                                                                   3!
                                                             C          *Eval. inside ( )
                                                            3  3
                                                                ( 3 3 3!
                                                                    )
                                                                     !
                                                                  
                                                             3!
                                                             !
                                                            0 3!         *Cancel out 3!'s
                                                             3!
                                                            ( 1 3!
                                                             )
                                                             1
                                       If you contain a factorial key, you can place it in as 3!, divided by 0!, divided by 3! and then
                                       press enter or =.
                                       If you don’t contain a factorial key, you can abridge it as exposed above and then enter it
                                       in.  It is perhaps best to shorten it first, since in some cases the numbers can get somewhat
                                       large, and it would be awkward to multiply all  those numbers one by one.
                                       This shows that there is only 1 draw out of the 56 established in part a that would enclose
                                       3 RED marbles.
                                   (c)  How many different draws would contain 1 red and 2 white marbles?
                                       This would be a combination problem, since a draw would be a group of marbles without
                                       regard to order.  It is similar to grabbing a handful of marbles and gazing at them.





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