A professor writes 40 discrete mathematics true/false questions. Of the statements in these questions, 17 are true. If the questions can be positioned in any order, how many different answer keys are possible?

/ Permutations and combinations / By

A powerful tool for solving problems is to transform a given problem into another one that is easier to solve, or that we already know how to solve. Then, from the solution to that new problem, we can give the solution to the former one.

This is such a case.

We know that we have 40 elements in total (40 true/false questions). We also know that 17 questions must be answered as true.

Let’s assume that 1 way that questions can be positioned is to have the 17 that are true first. In other words, questions 1 from 17 are true. The rest of the questions are false.

If we consider the index of the questions (1 to 17 and 18 to 40), we can see that they form a 17-combination of the first 40 integer numbers.

Why a combination and not a permutation?

As we are considering keys to answers (and not the questions), the order does not matter.

Let’s see a reduced example.

We have three questions. Q1 and Q2 are true, and Q3 is false.

We can order them as follows:

  1. Q1, Q2, Q3.  TTF
  2. Q1, Q3, Q2. TFT
  3. Q2, Q1, Q3. TTF
  4. Q2, Q3, Q1. TFT
  5. Q3, Q1, Q2. FTT
  6. Q3, Q2, Q1. FTT

Notice how key 1 and 3 are the same. Also, 2 and 4 are the same.

So, to find the answer we must calculate C(40,17).

C(40,17)= 40!/17!(40-17!)=40!/17!23!= 88732378800

Answer:

There are 88732378800 possible different answer keys.

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