in an examination of 9 papers, a candidate has to pass in more papers than the number of papers (contd)?

Question

in which he fails, in order to be successful. The number of ways in which he can be unsuccessful (remember its unsuccessful) are _________ ?
answer-256
if the calculation is too difficult to be typed on the computer, then plz indicate the steps in words that will lead me to the answer, though i will appreciate if u could give a full answer. thanks a lot for helping me out.

Answer 1

Hi,

Each test has 2 outcomes - either passed or failed, so there are 9 tests that will have 1 of 2 results. So long as the candidate passes more than he fails he will be successful, so he would be "successful" if he passed 5 of 9, 6 of 9, 7 of 9, 8 of 9 or all 9 of the papers. Since it does not matter which of the 9 papers he passes as long as he gets enough, you are interested in finding how many different ways he could pass enough of the 9 papers to be successful. This requires finding combinations of 9 papers taken 5,6,7,8,or 9 at a time. This can be done on most calculators using the nCr command. If you find that 9nCr9 = 1, that says there is only one way to pass all 9 of 9 games. 9nCr8 = 9, which says that there are 9 ways to pass 8 of the 9 games, since you could obviously fail any one of the 9 different games and still pass the required 8 games. 9nCr7 = 36, so there are 36 different ways to pass 7 of 9 papers. 9nCr6 = 84, so there are 84 different ways to pass 6 of the 9 total papers. Finally, 9nCr5 = 126, so there are 126 different ways to pass 5 of the 9 papers. Adding these numbers all together, 1 + 9 + 36 + 84 + 126 = 256.

I hope that helps.

Answer 2

The number of ways he can fail is what we want to know.
He can fail if he fails exactly 5 papers, or 6 or 7 or 8 or 9.


There is only 1 way to fail 9 exactly papers.
There are 9 ways to fail 8 exactly papers (he gets oly one right, and he can do that 9 ways).
To fail 7 papers, we find out the number of ways to pass eactly two: the two we can pass can be 1 and 2, 1 and 3,.... 1 and 9, 2 and 3, 2 and 4.., 2 and 9, 3 and 1...
this is called sampling without replacement and the number of ways to do this is n!/(n-k)! which is 9!/((9-2)!2!) = 9*8/2!.
So, to fail exactly 6 papers, we have 9!/(9-3)!3! = 9*8*7
To fail 5 papers, we have 9*8*7*6/4!
So, add all of these together, and we have 256! I used excel to do this!

Answer 3

Both previous solutions are correct, but here is a completely different way. Note that for each paper there are two possible outcomes. So the possible number of outcomes for all papers is: 2^9 = 512. Now notice that if a student passed with a particular outcome, then if he had reversed his performance on each paper (failed as opposed to passed, passed as opposed to failed), then he would have failed. So you can see that the number of ways to be successful is the same as the number of ways to unsucessful. So since the two add up to 512, the number of ways to be unsuccessful is 512/2 = 256.