So question2....
There are 12 questions on an examination, and each student must answer 8 questions including at least 4 of the first 5 questions. how many different combinations of questions could a student choose to answer?
There are two ways to do what's required:
(1) Answer all 5 of the first 5 questions, and 3 of the remaining 7.
(2) Answer 4 of the first 5 five questions, and 4 of the remaining 7.
For (1), there's only one way to answer 5 of 5, and 7 choose 3 = 35 ways to select 3 from 7. So there are 35 ways to accomplish (1).
For (2), there are 5 ways to select 4 of the first 5, and 7 choose 4 = 35 ways to select 4 of the remaining 7. So there are 5 * 35 = 175 ways to accomplish (2).
In total, then, I believe the answer would be 35+175=210 ways to do either (1) or (2).
- Steve J
Hello, Jason2JZ!
We must break this down into two cases . . .There are 12 questions on an examination.
Each student must answer 8 questions including at least 4 of the first 5 questions.
How many different combinations of questions could a student choose to answer?
[1] The student answers exactly 4 of the first 5 questions.
. . .He has $\displaystyle {5\choose4} = 5$ choices.
Then he picks 4 more questions from the last 7 questions.
. . .He has $\displaystyle {7\choose4} = 35$ choices.
He has: $\displaystyle 5 \times 35 \:=\:175$ ways if he picks 4 of the first 5 questions.
[2] The student answers all 5 of the first 5 questions.
. . .He has 1 way to do this.
The he picks 3 more question from the last 7 questions.
. . .He has $\displaystyle {7\choose3} = 35$ choices.
He has: $\displaystyle 1 \times 35 \:=\:35$ ways if he picks all 5 of the first 5 questions.
Therefore, he has a total of: $\displaystyle 175 + 35 \:=\:\boxed{210}$ ways to choose his 8 questons.