Question

In an examination of 9 papers, a candidate has to pass in more papers than the number of papers in
which he fails in order to get the success. The number of ways in which he can fail

Answer 1

pass : fail
4 : 5
3 : 6
2 : 7
1 : 8
0 : 9

Answer 2

sorry. But i think we've 2 use nPr formula here. I actually dont know how 2 use that one. Can u help me?

Answer 3

sorry, i didn't learnt that yet.

Answer 4

If he fails 5 papers out of the nine he fails. How many ways can he fail 5 papers? "9 choose 5" = 9C5 = 9!/5!(9-5)! = 9!/5!4! = 126

He can also fail 6 papers, or 7 or 8 or 9. Find the number he can fail each of those and them together to the number above.

Answer 5

sorry, i had made a mistake. the correct answer will be 256 ways!

Answer 6

Yes exactly, as explained above, the number of ways is:
9C5 + 9C6 + 9C7 + 9C8 + 9C9
= 126 + 84 + 36 + 9 + 1
= 256