A question paper has 5 questions. Each question has an alternative. The number of ways in which a student can attempt at least one question is
(a) \[{{2}^{5}}-1\]
(b) \[{{3}^{5}}-1\]
(c) \[{{3}^{4}}-1\]
(d) \[{{2}^{4}}-1\]
Answer
641.1k+ views
Hint: We solve this problem by using the combinations. We find the number of ways of attempting one question by selecting 1 question from 5 questions. The formula for selecting \['r'\] questions from \['n'\] questions is given as
\[\Rightarrow N\left( r \right)={}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
By using the above formula we find the number of ways of attempting at least one question.
Complete step by step solution:
We are given that there are 5 questions.
We are asked to find the number of ways of attempting at least 1 question.
Let us assume that number of ways of attempting at least 1 question as \['N'\]
We know that at least 1 means that 1 or more than 1
Let us assume that number of ways of attempting \['x'\] questions as \[N\left( x \right)\]
Now, we know that the number of ways of attempting at least 1 question is given as
\[\Rightarrow N=N\left( 1 \right)+N\left( 2 \right)+N\left( 3 \right)+N\left( 4 \right)+N\left( 5 \right)......\] equation(i)
We know that the attempting question is nothing but selecting the question from the 5 questions.
We know that the formula for selecting \['r'\] questions from \['n'\] questions is given as
\[\Rightarrow N\left( r \right)={}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
By using the above formula to equation (i) we get
\[\begin{align}
& \Rightarrow N={}^{5}{{C}_{1}}+{}^{5}{{C}_{2}}+{}^{5}{{C}_{3}}+{}^{5}{{C}_{4}}+{}^{5}{{C}_{5}} \\
& \Rightarrow N=5+10+10+5+1 \\
& \Rightarrow N=31 \\
\end{align}\]
Let us rewrite the number 31 in such a way that we can get in the form \[{{2}^{5}}\] then we get
\[\begin{align}
& \Rightarrow N=32-1 \\
& \Rightarrow N={{2}^{5}}-1 \\
\end{align}\]
Therefore, the number of ways of attempting at least 1 question is \[{{2}^{5}}-1\]
So, option (a) is the correct answer.
Note: This problem can be solved in another method.
Here, we have two choices of attempting a question that is either attempting or not attempting.
So, the number of ways of answering a question is 2
We know that the total number of ways attempting \['n'\] questions having \['r'\] possibilities for each question is given as
\[\Rightarrow N={{r}^{n}}\]
By using the above formula the number of ways of attempting 5 questions having 2 possibilities is given as
\[\Rightarrow N={{2}^{5}}\]
We are asked to find the number of ways of attempting at least 1 question
We get the required number of ways by subtracting the number of ways of not attempting all questions from the total number of ways
There is only 1 way of not attempting all questions.
Therefore, the number of ways of attempting at least 1 question is
\[\Rightarrow n={{2}^{5}}-1\]
So, option (a) is the correct answer.
\[\Rightarrow N\left( r \right)={}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
By using the above formula we find the number of ways of attempting at least one question.
Complete step by step solution:
We are given that there are 5 questions.
We are asked to find the number of ways of attempting at least 1 question.
Let us assume that number of ways of attempting at least 1 question as \['N'\]
We know that at least 1 means that 1 or more than 1
Let us assume that number of ways of attempting \['x'\] questions as \[N\left( x \right)\]
Now, we know that the number of ways of attempting at least 1 question is given as
\[\Rightarrow N=N\left( 1 \right)+N\left( 2 \right)+N\left( 3 \right)+N\left( 4 \right)+N\left( 5 \right)......\] equation(i)
We know that the attempting question is nothing but selecting the question from the 5 questions.
We know that the formula for selecting \['r'\] questions from \['n'\] questions is given as
\[\Rightarrow N\left( r \right)={}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
By using the above formula to equation (i) we get
\[\begin{align}
& \Rightarrow N={}^{5}{{C}_{1}}+{}^{5}{{C}_{2}}+{}^{5}{{C}_{3}}+{}^{5}{{C}_{4}}+{}^{5}{{C}_{5}} \\
& \Rightarrow N=5+10+10+5+1 \\
& \Rightarrow N=31 \\
\end{align}\]
Let us rewrite the number 31 in such a way that we can get in the form \[{{2}^{5}}\] then we get
\[\begin{align}
& \Rightarrow N=32-1 \\
& \Rightarrow N={{2}^{5}}-1 \\
\end{align}\]
Therefore, the number of ways of attempting at least 1 question is \[{{2}^{5}}-1\]
So, option (a) is the correct answer.
Note: This problem can be solved in another method.
Here, we have two choices of attempting a question that is either attempting or not attempting.
So, the number of ways of answering a question is 2
We know that the total number of ways attempting \['n'\] questions having \['r'\] possibilities for each question is given as
\[\Rightarrow N={{r}^{n}}\]
By using the above formula the number of ways of attempting 5 questions having 2 possibilities is given as
\[\Rightarrow N={{2}^{5}}\]
We are asked to find the number of ways of attempting at least 1 question
We get the required number of ways by subtracting the number of ways of not attempting all questions from the total number of ways
There is only 1 way of not attempting all questions.
Therefore, the number of ways of attempting at least 1 question is
\[\Rightarrow n={{2}^{5}}-1\]
So, option (a) is the correct answer.
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