Question

A student has to answer \[10\] questions, choosing at least $4$ questions from each of parts $A$ and $B$. If there are $6$ questions in part $A$ and $7$ questions in part $B$, in how many ways can the student choose \[10\] questions?

Answer 1

Hint: First we count the total number of questions and then read the conditions given in the question. A student has to choose at least $4$ questions from each of parts $A$ and $B$, so we will find all possible ways to choose questions with this condition.
As we know that the formula for combination is \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
Where, $n=$ number of items/objects
And $r=$ number of items/objects being chosen at a time
Here in this question we use the above formula.

Complete step-by-step solution -
We have given that total number of questions in part $A=6$
Also, we have given that total number of questions in part $B=7$
So, the total number of questions will be $=13$
Also, we have given that a student has to answer \[10\] questions out of $13$ questions.
A student has to choose at least $4$ questions from each of parts $A$ and $B$. It means minimum $4$ questions must be chosen from a part, not less than $4$ questions. So, there are following possible ways to choose \[10\] questions –
First one is that a student can choose $4$ questions from part $A$ and $6$ questions from part $B$.
Second one is that a student can choose $5$ questions from each of parts $A$ and $B$.
Third one is that a student can choose $6$ questions from part $A$ and $4$questions from part $B$.
So, when we combine all possible ways the total number of ways a student can choose \[10\] questions will be- \[\left( {}^{6}{{C}_{4}}\times {}^{7}{{C}_{6}} \right)+\left( {}^{6}{{C}_{5}}\times {}^{7}{{C}_{5}} \right)+\left( {}^{6}{{C}_{6}}\times {}^{7}{{C}_{4}} \right)\]
Now, by using the formula of combination \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
The total number of ways a student can choose \[10\] questions will be
 $\begin{align}
  & =\left( 15\times 7 \right)+\left( 6\times 21 \right)+\left( 1\times 35 \right) \\
 & =105+126+35 \\
 & =266 \\
\end{align}$
So the total number of ways a student can choose \[10\] questions will be $266$.

Note: When the word ‘at least’ is given in the question, most of the students got confused. The word ‘at least’ means that there is a restriction on minimum selections. In the given question a student has to choose at least 4 questions from each of parts A and B , it means a student can never choose less than 4 questions from any part but a student can choose more than 4 questions from any part.