Question

There are $12$ true-false questions in an examination. How many sequences of answers are possible?

Answer 1

Hint: We will solve this question using the simple selection method without using the combination formula to find the total number of sequences of answers possible in a$12$ true and false questions in an examination.

Complete step by step Answer :
We have been given the number of questions which is $12$. These are true and false questions which means that a student can answer a single question with either true or false.
Therefore, the number of ways in which each question can be answered is $2$ ways.
Since there are $12$ questions, and each question can be answered in $2$ ways then:
1st question can be answered in $2$ ways.
2nd question can be answered in $2$ ways.
3rd question can be answered in $2$ ways.
And so on till $12$ questions.
The $2$questions can be answered in $2 \times 2$ ways.
The $3$questions can be answered in $2 \times 2 \times 2$ ways.
The $4$questions can be answered in $2 \times 2 \times 2 \times 2$ ways and so on.
So, after solving all the $12$ questions, we get

$$2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \\\ = {2^{12}} \\\ = 4096 \\\$$

Therefore, there are $4096$ sequences of answers possible.

Note: We can also solve this question using the combination formula $\dfrac{{n!}}{{r!\left( {n - r} \right)!}}$. We will either select true or false to answer a question, so the total number of possibilities is $2$ and the probable outcome is$1$. Hence, we can say that the number of ways each question would be answered is ${}^2{C_1}$. There are $12$ questions so ${}^2{C_1}$ is multiplied $12$ times giving us ${({}^2{C_1})^{12}} = {\left( {\dfrac{{2!}}{{1!\left( {2 - 1} \right)!}}} \right)^{12}} = {\left( {\dfrac{{2!}}{{1!1!}}} \right)^{12}} = {2^{12}} = 4096$ ways.