Question

An examination, a question paper consists of 12 questions divided into two parts i.e. part 1 and part 2, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions?

Answer 1

Hint: In this question, we need to list all the combinations possible for example if we choose 3 questions in the first part then we need to choose 5 questions from the part 2. Then we need to find the possible ways for each of the combinations listed using the combinations formula given by ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$. Now, all the ways obtained from different combinations to get the result.

Complete step by step solution:
COMBINATIONS:
Each of the different groups or selections which can be made by some or all of a number of given things without reference to the order of the things in each group is called a combination.
The number of combinations of n different things taken r at a time is
${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$, $0\le r\le n$
Now, we need to do 8 questions in which we need to choose at least 3 out of 5 questions and 7 questions in each section
Let us now list all the possible combinations possible to select 8 question with at least 3 from part 1, part 2
If we choose 3 from part 1 then we need to choose 5 from part 2
If we choose 4 from part 1 then we need to choose 4 from part 2
If we choose 5 from part 1 then we need to choose 3 from part 2
Now, the number of possible ways of choosing 3 from part 1 and 5 from part 2 is given by
$\Rightarrow {}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$
Here, there are 5 questions in part 1 in which we need to choose 3 and 7 questions in part 2 out of which we need to choose 5
$\Rightarrow {}^{5}{{C}_{3}}\times {}^{7}{{C}_{5}}$
Now, this can be further written as
$\Rightarrow \dfrac{5!}{3!2!}\times \dfrac{7!}{5!2!}$
Now, on cancelling out the common terms we get,
$\Rightarrow \dfrac{7\times 6\times 5\times 4\times 3\times 2}{3\times 2\times 2\times 2}$
Now, on further simplification we get,
$\Rightarrow 210$
Now, the number of ways of choosing 4 questions from part 1 and 2 is given by
$\Rightarrow {}^{5}{{C}_{4}}\times {}^{7}{{C}_{4}}$
Now, this can be further written as
$\Rightarrow \dfrac{5!}{4!1!}\times \dfrac{7!}{4!3!}$
Now, on cancelling out the common terms we get,
$\Rightarrow \dfrac{5\times 7\times 6\times 5\times 4\times 3\times 2}{4\times 3\times 2\times 3\times 2}$
Now, on further simplification we get,
$\Rightarrow 175$
Now, the number of ways of choosing 5 questions from part 1 and 3 questions from part 2 is given by
$\Rightarrow {}^{5}{{C}_{5}}\times {}^{7}{{C}_{3}}$
Now, this can be further written as
$\Rightarrow \dfrac{5!}{5!1!}\times \dfrac{7!}{4!3!}$
Now, on cancelling out the common terms we get,
$\Rightarrow \dfrac{7\times 6\times 5}{3\times 2}$
Now, on further simplification we get,
$\Rightarrow 35$
Let us now calculate the total number of ways possible
$\Rightarrow 210+175+35$
Now, on further simplification we get,
$\Rightarrow 420$
Hence , there are 420 ways possible to choose 8 questions from 5 questions of part 1 and 7 questions of part 2 with at least 3 from each.

Note: Instead of finding the number of ways for each combination and then adding them to get the total number of ways we can directly simplify it by considering all the combinations at a time and simplify accordingly. This method reduces the number of steps.
It is important to note that there will be different combinations possible as there is a given condition that there should be at least 3 questions from each of the parts which means 3 or more. It is also to be noted that we cannot choose more than 5 questions from part 1 as there are only 5 questions.