a Given that a test has 10 questions, we are asked to find how many ways we can choose 4 of the first 6 questions. Note that as long as we select 4 of the first 6 questions, the order does not matter. We can solve this problem by using combinations. The number of combinations of n objects chosen r at a time is given as follows.
_nC_r = n!/(n-r)!r!
Now, let's evaluate the formula when n= 6 and r= 4.
Therefore, there are 15 different ways of choosing 4 questions from the first 6 questions on the test.
b We are told that we must answer 7 questions from the test, including exactly 4 of the first 6 questions. Given that we already answered 4 of the first 6 questions, the other 2 questions from the first 6 are not available anymore. Therefore, the remaining questions are given by the difference of the total questions 10 and 6.
10- 6= 4questions left
Therefore, there are 4 questions left. Now, to get to 7 answered questions we still need to answer 3 questions of the remaining ones.
c From the 4 remaining questions we have to answer 3 of them to finish the test. Again, the order we choose them in is not important. Therefore, we can use combinations to find how many ways we can choose the 3 questions from the 4 available.
Therefore, there are 4 different ways of choosing 3 questions from the 4 remaining questions.
d We have found that there are 15 different ways to select the first 4 questions and 4 ways for the last 3 ones. Therefore, we can use the Multiplication Counting Principle to find the number of different ways of completing the test and meeting the requirements. Let's do it!
