SOLVE THIS QUESTION !!! i'll give you 10 pts !! please show the work !?

Question

An examination consists of 13 questions. A student must answer only one of the 1st 2 questions and only nine of the remaining ones. How many choices does the student have?

i dont need the answer!!! I NEED THE WORK

Answer 1

For the first question, the student has 2 choices (obvious).

Next, the student must skip two of the remaining 11 questions. To pick these two, he has

(11 . 10)/2 = 55

options. In total, therefore, there are

2 . 55 = 110

choices.

Answer 2

110 ways:

There are 2 ways to choose 1 of the first 2 questions
2C1 = 2!/(2-1)!(1!) = 2

There are 55 ways to choose 9 of the remaining 11 questions.
11C9 = 11!/(11-9)!(9!) = (11*10)/2 = 55

Another way to think of this... how many ways to pick the first question? Obviously you pick 1, or you pick 2. So 2 ways.

Now how many ways to pick 9 of the 11 remaining? Well, an easier question is how do you pick 2 to *skip*? There are 11 choices for the first question to skip, 10 choices for the second to skip. That would be 110, except that we don't care about order. For example if I skip #3 and then #7, that's the same as if I skipped #7 and then #3. Since half the combinations are duplicates, we divide by 2. So the real answer is 110/2 or 55.

Multiplying these together we have 2 ways to pick among the first 2 questions, and 55 ways to pick among the remaining 11.

= 2 x (11 x 10)/2
= 2 x 55
= 110

So the answer is 110.

Answer 3

Doing the maths bit, you have a choice of 2 out of the first 2 questions to answer ... then you have 9 questions to answer out of the remaining 11

Hence the textbook would probably say: 2 X C11/9 (hmmm - how to write the combination notation?)

2 X 11 X 10 X 9 X 8 X 7 X 6 X 5 X 4 X 3 choices (close enough to 11! anyway).

That is to say, 11 possibilities for your 1st question, 10 remaining for your 2nd, 9 for your 3rd etc...

In actual fact, there are more choices in real life. You could do ignore the exam instructions and do any or none of the questions. You could log onto yahoo. You could go surfing!

Hope this helps.

Answer 4

3

Answer 5

11

Answer 6

only 2 people so far seem to know what they are doing!
bandf seems to have the clearest explanation thus far for the correct answer of 110.

just to expand a bit of the explanation of why there are 55 ways of choosing 9 of 11 questions. bandf was right on in saying this is the same as choosing the 2 questions NOT to answer. So, why is that 55? As bandf pointed out, there are 11 ways to choose the first question not to answer, and then 10 remaining questions from which to choose the second question not to answer. What I wanted to try to clarify is why the answer 11 x 10 is cut in half. In the above explanation, choosing first not to answer question 5, then choosing not to answer question 7 is counted as different from first choosing not to answer question 7, then choosing not to answer question 5. In reality, these are the same, not different. Thus, 11 x 10 double counts the possibilities, leaving 55.

Answer 7

Question 1
Question 2 Answer

Question 3-13 Answer only 9 of the remaining 11


I was never good at this ..... LOL
13 choices as the student has the choice to answer it or not.

Answer 8

AHA!! This is a trick question.

The student has 2 choices. He/she must (1) choose which of the first two questions to answer. Then he/she must (2) choose 9 more of the remaining 11 questions.

Answer 9

1 of the first 2, you use combinations, like 2C1, and then 9 of the last 11, 11C9. The formula for nCr is
n!
____
(n-r)!r!

so the answer is
11! 2!
----- * --------
9! * 2! 1!* 1!

The ! means factorial, like 9! = 9*8*7*6*5*4*3*2*1

so the answer is (11*10*9*8*7*6*5*4*3*2*1)/(9*8*7*6*5*4*3*2*1*2*1) * (2*1) / 1*1

it cancels out to 11*10/2*2 = 110

Answer 10

I must be missing something it seems too easy. 1 of the first two, then any 9 of the remaining 11. That leaves 220 different combinations of answering. Sorry I cant show the math on my computer.