November problem of the month help!!!

Question

Megan has four discount coupons that she can use. Values are $8,$13,$17,$22. How many different amounts can she save by using one or more of these four coupons? 1) show all the combinations you find? 2) how may different amounts possible?

Answer 1

A "problem of the month" should be read very carefully.

 

There are 4 coupons.  The choice is whether to use or not to use each one with the constraint of "using one or more."  To count in binary with four digits we go from 0 (not allowed) to 31 (that is, 1111).  These are all the combinations (PLZ don't simply find the _nC_r values) and the corresponding amounts saved:

 

 22  17   13   8   saved      [PLZ note:  high to low]

  0    0     0    1      8          [ 0000 not allowed;  this is binary 0001 ]

  0    0     1    0     13             [0010 in binary is 2]

  0    0     1    1     21

  0    1     0    0     17

  0    1     0    1     25

  0    1     1    0     30

  0    1     1    1     38

  1    0     0    0     22

  1    0     0    1     30

  1    0     1    0     35

  1    0     1    1     43

  1    1     0    0     39

  1    1     0    1     47

  1    1     1    0     52

  1    1     1    1     60        [this is decimal 15 in binary]

 

There are 15 combinations.  However, there are 14 "different amounts" (PLZ find duplicate.

 

Special note:  One way to make it more difficult to correctly answer a problem is to ask a question that is not quite the usual question asked.  For example, "A bus driver started the route, picked up 2 people at the first stop, picked up 5 people at the next stop, picked up 3 people at the next stop but 2 got off, picked up 1 person at the next stop but 2 got off, . . . , and let off the remaining people at the last stop.  How many stops did the bus make?"

Answer 2

They threw in a "gotcha".

8+22 = 13+17 = 30

This is the only possible duplication.

After you find all of the combinations, you have to subtract 1 to get the total number of different values.

 

Answer 3

₄C₁ + ₄C₂ + ₄C₃ + ₄C₄

 

Or make a table

8, 13, 17, 22

8+13, 8+17, 8+22, 13+17, 13+22, 17+22

8+13+17, 8+13+22, 8+17+22, 

 

And continue