Choosy Examiner

Probability Level 3

A candidate is required to answer 6 out of 10 questions which are divided into two groups each containing five questions and he is not permitted to attempt more than 4 from each group. In how many ways can he make up his choice?

For part 1 try this .

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The answer is 200.

2 solutions

Discussions for this problem are now closed

Melissa Quail
Feb 1, 2015

Ignoring the condition that no more than four questions can be attempted from each section, there are 10C6 ways of choosing 6 questions to answer out of the 10 questions.

Then, of these, 10 ways involve answering more than 4 questions from one of the sections:

Either all questions from section 1 and one of the five questions from section 2 (five ways)

Or all the questions from section 2 and one of the five questions from section 1 (five ways), giving a total of 10 ways that do not fit the criteria.

Subtract this from the total to give

10C6 - 10= $\boxed{200}$ ways.

Majed Khalaf
Feb 3, 2015

To be able to answer this question, we take the possibilities and go on according to them: The first possibility is: 4 questions from the first group and 2 questions from the second one, and vice versa, or three from the first group and three from the second group. We calculate the number of choices of each possibility and add them up, ie: 5 4 3/3 2=5, 5 5 4/2=50 , 50 is half of the choices of possibility one, so we double the amount 50 2=>100, and about the other possibility: 5 4 3/3 2=10, 10 10=100, and then we add them up, 100+100=200 choices, the reason behind not doubling the second possibility is because they're equal in the count of numbers taken.