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Question: A question paper has 5 questions. Each question has an alternative. The number of ways in which a st...

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) 25−1{{2}^{5}}-1
(b) 35−1{{3}^{5}}-1
(c) 34−1{{3}^{4}}-1
(d) 24−1{{2}^{4}}-1

Explanation

Solution

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′'r' questions from ′n′'n' questions is given as
⇒N(r)=nCr=n!r!(n−r)!\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′'N'
We know that at least 1 means that 1 or more than 1
Let us assume that number of ways of attempting ′x′'x' questions as N(x)N\left( x \right)
Now, we know that the number of ways of attempting at least 1 question is given as
⇒N=N(1)+N(2)+N(3)+N(4)+N(5)......\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′'r' questions from ′n′'n' questions is given as
⇒N(r)=nCr=n!r!(n−r)!\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

& \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{aligned}$$ Let us rewrite the number 31 in such a way that we can get in the form $${{2}^{5}}$$ then we get $$\begin{aligned} & \Rightarrow N=32-1 \\\ & \Rightarrow N={{2}^{5}}-1 \\\ \end{aligned}$$ Therefore, the number of ways of attempting at least 1 question is $${{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.