Question

A candidate is required to answer 6 out of 12 questions which are divided into two parts A and B, each containing 6 questions and he/she is not permitted to attempt more than 4 questions from any part. In how many different ways can he/she make up his/her choice of 6 questions? \[\]
A.850 \[\]
B.800\[\]
C.750\[\]
D. 700\[\]

Answer 1

Hint: We see that he/she can choose 6 questions under given condition in the following one of three cases. The first case is she/she chooses 4 questions from part A and 2 questions from part B, the second is she/she chooses 3 questions from part A and 3 questions from part B, the third case is she/she chooses 2 questions from part A and 4 questions from part B. We find number of ways for each case using the formula for selection of $r$distinct things from $n$ distinct things ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and the rule of proud. We add the number of ways obtained in each case to get the answer. \[\]

Complete step-by-step answer:
We are asked to find the total number of ways he/she makes up his/her choice of 6 questions with the condition that he/she can choose at most 4 questions from either part A or B. We have three cases here \[\]
Case-1: He/she chooses to answer 4 out the required 6 questions in part A and rest $6-4=2$ questions in part B. He/she can select 4 questions from available 6 questions in part A in ${}^{6}{{C}_{4}}$ ways and then he/she can also select 2 questions from available 6 questions in part B in $^{6}{{C}_{2}}$. We use rule of product and find the number ways he/she can choose to answer 6 question in case-1 as
\[{{N}_{1}}={}^{6}{{C}_{4}}\times {}^{6}{{C}_{2}}\]
Case-2: He/she chooses to answer 3 out the required 3 questions in part A and rest $6-3=3$ questions in part B. He/she can select 3 questions from available 6 questions in part A in ${}^{6}{{C}_{3}}$ ways and then he/she can also select 3 questions from available 6 questions in part B in $^{6}{{C}_{3}}$. We use rule of product and find the number ways he/she can choose to answer 6 questions in case-2 as
\[{{N}_{2}}={}^{6}{{C}_{3}}\times {}^{6}{{C}_{3}}\]
Case-3: He/she chooses to answer 2 out the required 6 questions in part A and rest $6-4=2$ questions in part B. He/she can select 2 questions from available 6 questions in part A in ${}^{6}{{C}_{2}}$ ways and then he/she can also select 4 questions from available 6 questions in part B in $^{6}{{C}_{4}}$. We use rule of product and find the number ways he/she can choose to answer 6 questions in case-3 as
\[{{N}_{3}}={}^{6}{{C}_{4}}\times {}^{6}{{C}_{2}}\]
We see that he/she can choose 6 questions following either of case-1, case-2 or case-3. We use rule of sum and find the total number of ways he/she choose 6 questions as
\[\begin{align}
  & {{N}_{1}}+{{N}_{2}}+{{N}_{3}}={}^{6}{{C}_{4}}\times {}^{6}{{C}_{2}}+{}^{6}{{C}_{3}}\times {}^{6}{{C}_{3}}+{}^{6}{{C}_{2}}\times {}^{6}{{C}_{4}} \\
 & \Rightarrow {{N}_{1}}+{{N}_{2}}+{{N}_{3}}=\dfrac{6!}{4!2!}\times \dfrac{6!}{2!4!}+\dfrac{6!}{3!3!}\times \dfrac{6!}{3!3!}+\dfrac{6!}{2!4!}\times \dfrac{6!}{4!2!} \\
 & \Rightarrow {{N}_{1}}+{{N}_{2}}+{{N}_{3}}=15\times 15+20\times 20+15\times 15 \\
 & \Rightarrow {{N}_{1}}+{{N}_{2}}+{{N}_{3}}=225+400+225=850 \\
\end{align}\]

So, the correct answer is “Option A”.

Note: We note from the fundamental of counting that if there are $m$ ways to do something and $n$ ways to another thing then by rule of product there are $m\times n$ ways to do both things simultaneously and by rule of sum $m+n$ ways do either of the things. We can alternatively solve using negation where we find total number of ways he/she can answer 6 questions as $^{12}{{C}_{6}}$ and then the number of ways he/she can answer more than 4 questions from either part as $2\left( {}^{6}{{C}_{5}}\times {}^{6}{{C}_{1}}+{}^{6}{{C}_{6}}\times {}^{6}{{C}_{0}} \right)$and then subtract from $^{12}{{C}_{6}}$.