Question

For a certain probability test, the teacher created 6 problems with different versions. For problems 1, 3, and 4, he created five versions of each. For problems 5 and 6 there are four versions of each, while for problem 2 there is only one version. How many different tests can be done if each test will have a version of each problem? If in addition to the different versions for each problem, the teacher decides to put the problems in random order, how many different tests are possible?

Answer 1

a) Given that for first part order is not important, the total number of tests possible here is computed using the product rule as:

= Number of different versions for problem 1 * Number of different versions for problem 2 * ..... Number of different versions for Problem 6

= 5³*4²*1

= 2000

Therefore there are a total of 2000 different tests possible based on only the selection of different versions of the 6 problems

b) Given that order of 6 problems selected is also taken here, the total number of tests here is computed as:

= 2000* Number of permutations of 6 problems selected

= 2000*6!

= 1440000

Therefore there are 1440000 different tests possible here.