Permutations and combinations

Probability Level 3

An examination paper has two parts A and B. There are six questions in part A and eight questions in part B. A candidate is required to do four questions in part A which must include at least one of question 1 or question 2, and any 3 from part B. In how many ways can the candidate complete the paper?


The answer is 784.

Chimere Udonsi by Chimere Udonsi , 18, Nigeria

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2 solutions

Chimere Udonsi
Dec 30, 2016

Since in part A, we are required to answer four questions we would have originally done 6 combination 4. But because we are also expected to answer either question 1 or 2, we now do 5 combination 3 multiplied by 2! (2 factorial). 5 combination 3 is 10 multiplied by 2(which is 2 factorial). Then multiply this result by 56(which is 8 combination 3-for part B). 20 multiplied by 56 gives the answer which is 1120.

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Can you elaborate on the approach of "5 combination 3, multiplied by 2"? What is the reasoning that allows us to do so?

I believe it is incorrect, and you are double counting various cases. In particular, note that there are at most ( 6 4 ) = 15 { 6 \choose 4 } = 15 ways to do 4 questions without restriction.

Calvin Lin Staff - 4 years, 5 months ago
Saya Suka
Apr 4, 2021

Answer
= [ # of choices for part A ] × [ # of choices for part B ]
= [ 6C4 – 4C4 ] × [ 8C3 ]
= ( 15 – 1 ) × ( 56 )
= 14 × 56
= 784

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