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?

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- Dec 11th 2008, 07:42 PMJason2JZOrganized counting
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? - Dec 11th 2008, 07:55 PMSteve_JOrganized counting
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 - Dec 11th 2008, 07:58 PMJason2JZ
ahh ic thank you!(Rock)

- Dec 11th 2008, 08:06 PMSoroban
Hello, Jason2JZ!

Quote:

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.