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.