SOLUTION: Suppose an exam has 28 questions on it and students need to answer 18 of the questions. How many ways can this be done if the first 5 and last 4 questions must be answered and que

Question 236081: Suppose an exam has 28 questions on it and students need to answer 18 of the questions. How many ways can this be done if the first 5 and last 4 questions must be answered and question #19 must not be answered because of a typo?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
28 questions
answer 18
must answer first 5 and last 4
must not answer number 19

numbers are 1 through 28
take away first 5 to get 6 through 28
take away last 4 to get 6 through 24
take away 19 to get:
6 through 18 and 20 through 24

18 have to be answered.

4 + 5 = 9 slots that have to be picked in all cases.

that leaves 9 slots that can be chosen out of the ones that are still available.

slots available for those 9 are 6 through 18 and 20 through 24

that's 13 + 5 = 18 slots that are available for the 9 slots that still have to be filled.

since 9 have to be answered, we have number of possible combinations of:

9 out of 18.

formula for number of combinations equals n! / ((n-x)!*x!)

with n = 18 and x = 9, this formula becomes:

18! / ((18-9)!*9!)

that becomes 18! / (9!*9!) = 48620 different ways

your answer is:

48620 different ways that the questions can be answered ***********************

to see how this works, pick smaller numbers.

assume the following:

numbers are 1 through 10

first 2 and last 1 have to be picked.

slot number 6 is not available.

available slots become 3 through 5 and 7 through 9.

you have to answer 5 questions

number of available slots is 3 and 3 = 6

number of slots frozen are 3

total of 5 slots have to be filled leaving 2 slots with options.

we have 6! / ((6-2)! * 2!)

this becomes 6! / (4! * 2!)

this becomes 6*5*4!/(4!*2!) which becomes 6*5/2*1 which becomes 3*5 = 15

the possible ways the test can be answered is 15.

let's see how that works out.

slots 1 and 2 and 10 are required.

slot 6 can't be used.

slots available are:

345789

how many possible combinations of 2 can we make of this?

we can make.............

34
35
37
38
39
45
47
48
49
57
58
59
78
79
89

that's a total of 15.

stick 12 in front of each of these and add 10 at the end of each of these and we have the full set including the frozen slots that had to be filled.

example substituting hexadecimal digit a for 10:

total of 5 slots have to be filled.

slots 1 and 2 and a are frozen for all possible combinations.

1234a
1235a
1237a
1238a
1239a
1245a
1247a
1248a
1249a
etc.....

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