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How many ways can a student do a ten-question true-false exam if he or she can choose not to answer any number of questions?

1 Answer

6 votes

Answer:

26,565

Explanation:

First, lets see how many different set of questions she can resolve. Well, se can solve only one question or she can solve 2 questions, or 3, or 4, or so on until 10 which is the maximum number of questions she can solve.

If she solves 1 question, say question number 5, she can put True (T) or False (F). So, we have two possibilities. However, she also can chose which question she will solve from the set of 10 questions. The way to account for all the options she has is using combinations. In this case the has combinations of 10 takes as 1: 10 C 1, which in this case is obviously 10 (it's easy, there are 10 ways of picking something from 10 options). So, is she answeres only 1 quesitons she has to chose which question it is from 10 options and then what she answers T or F, and the possibilities for 1 question (Q1) are:

Q1 = [10 C 1] * 2 = 20

Now, if she answers 2 questions, she has to pick such 2 questions from the set of 10: 10 C 2 = 45. After that she needs to chose the answers, and as every questions has 2 options she can answer in 2*2 ways, which is equal to 4. So for answering 2 questions (Q2):

Q2 = [10 C 2] * 2*2 = [10 C 2] * 2^2 = 180

Notice that for every amount of quesitons que can figure it the same way. For every number of questions x between 1 and 10, we can say that she has 10 C x number of ways of answering and after decided she has 2^x ways of answering. So for any number x:

Qx = [10 C x] *2^x

For the next cases we have:

Q3 = [10 C 3] *2^3 = 120 * 8 = 960

Q4 = [10 C 4] *2^4 = 3360

Q5 = [10 C 5] *2^5 = 8604

Q6 = [10 C 6] *2^6= 13440

Q7 = [10 C 7] *2^8 = 15360

Q8 = [10 C 8] *2^8 = 11520

Q9 = [10 C 9] *2^9 = 5120

Q10 = [10 C 10] *2^10 = 1024

Summing all:

Total = 26,564‬

We have to add 1 as there is also the possibility of answering no question at all. So, the total number of questions is 26,565

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