Question

A candidate is required to answer 6 questions by choosing at least one question from each section, where ${{1}^{st}}$ section consists of 5 questions, ${{2}^{nd}}$ section consists of 3 questions and ${{3}^{rd}}$ section consists of 2 questions. In how many ways can he make up his choice?
(a) 225
(b) 320
(c) 175
(d) 200

Answer 1

Hint: In order to solve this problem, we have to find the number of possible combinations of the question to be attempted from each section. Once it's decided then we need to take the combination to get all the possible ways. The formula for the number of possible ways the $r$ thing is don’t out of $n$ things is given by ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ . By adding all the possibilities, we will get the total ways to attempt the questions

Complete step by step answer:
There are a total of 3 sections consisting of 5, 3, 2 questions in each section respectively.
A student needs to answer at least 6 questions and also least one question can be selected from one section.
Therefore, all the possible combinations for attempting the questions are
411, 321, 312, 231, 222, 132.
Where the left the left-most digit shows the questions attempted from the first section, the second digit shows the number of questions attempted in the second section and the last digit shows the number of questions attempted in the last section.
We can see that the summation of all the digits always adds up to 6.
Considering the first case for the first section we need to attend 4 any 4 questions from a total of 5 questions so the number of ways it can be done can be shown by combination relation.
to find the probability of at least one bulb working. Also, we need to use the relation of combinations. The formula for the number of possible ways the $r$ thing is don’t out of $n$ things is given by ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ .
The number of ways of attending 4 questions out of 5 is given by ${}^{5}{{C}_{4}}$ .
Similarly, for the second section, the number of ways of solving 1 question out of 3 is given by ${}^{3}{{C}_{1}}$ .
The total number of ways by the first combination is $\left( {}^{5}{{C}_{4}}\times {}^{3}{{C}_{1}}\times {}^{2}{{C}_{1}} \right)$ .
Similarly following the same procedure for all the 6 combinations and then adding up all the probabilities we get,
The total no. of ways of attempting question papers is $\begin{align}
  & =\left( {}^{5}{{C}_{4}}\times {}^{3}{{C}_{1}}\times {}^{2}{{C}_{1}} \right)+\left( {}^{5}{{C}_{3}}\times {}^{3}{{C}_{2}}\times {}^{2}{{C}_{1}} \right)+\left( {}^{5}{{C}_{3}}\times {}^{3}{{C}_{1}}\times {}^{2}{{C}_{2}} \right)+\left( {}^{5}{{C}_{2}}\times {}^{3}{{C}_{3}}\times {}^{2}{{C}_{1}} \right)+\left( {}^{5}{{C}_{2}}\times {}^{3}{{C}_{2}}\times {}^{2}{{C}_{2}} \right) \\
 & +\left( {}^{5}{{C}_{1}}\times {}^{3}{{C}_{3}}\times {}^{2}{{C}_{2}} \right) \\
\end{align}$ .
Calculating, ${}^{5}{{C}_{4}}=\dfrac{5!}{4!\left( 5-4 \right)!}=\dfrac{5}{1}=5$ ,
${}^{3}{{C}_{1}}=\dfrac{3!}{1!\left( 3-1 \right)!}=3$ ,
${}^{2}{{C}_{1}}=\dfrac{2!}{1!\left( 2-1 \right)!}=2$ ,
${}^{5}{{C}_{3}}=\dfrac{5!}{3!\left( 5-3 \right)!}=10$ ,
${}^{3}{{C}_{3}}=\dfrac{3!}{3!\left( 3-3 \right)!}=1$ ,
${}^{5}{{C}_{2}}=\dfrac{5!}{2!\left( 5-2 \right)!}=10$ ,
${}^{3}{{C}_{2}}=\dfrac{3!}{2!\left( 3-2 \right)!}=3$ ,
${}^{2}{{C}_{2}}=\dfrac{2!}{2!\left( 2-2 \right)!}=1$ ,
${}^{5}{{C}_{1}}=\dfrac{5!}{1!\left( 5-1 \right)!}=\dfrac{5}{1}=5$ .
Substituting all these formulas we get,
The total no. of ways of attempting question papers is $=\left( 5\times 3\times 2 \right)+\left( 10\times 3\times 2 \right)+\left( 10\times 3\times 1 \right)+\left( 10\times 1\times 2 \right)+\left( 10\times 3\times 1 \right)+\left( 5\times 1\times 1 \right)=30+60+30+20+30+5=175$ .
Therefore, the number of ways of attending question papers is 175.
Hence, the correct option is (c).


Note:
The combination like 510 is not allowed as the first condition says that we need to attempt at least one question from each section. Also, the combination 141 is not allowed because in the section there are only three questions and there is no way we can attempt 4 questions. The probabilities get multiplied in between sections because all the events are performed simultaneously.