Question

A candidate is required to answer 6 questions by choosing at least one question from each section, where section 1 consists of 4 questions, section 2 consists of 3 questions and section 3 consists of 2 questions. In how many ways can he make a choice?
A.76
B.80
C.95
D.63

Answer 1

The given question can be solved using combinations of choosing 6 questions, at least 1 from each. We can get the correct answer by finding the six possible combinations and by adding it. The definition of ‘combination’ is selection of items from a collection, such that the order of selection does not matter (unlike permutations).

Complete step-by-step answer:
A candidate has to choose 6 questions from 3 sections, at least 1 from each.
The first way of choosing 6 questions can be 4 from section 1, 1 from section 2 and 1 from section 3. This can be written as ${}^4{C_4} \cdot {}^3{C_1} \cdot {}^2{C_1}$ .
The second way of choosing 6 questions can be 3 from section 1, 2 from section 2 and 1 from section 3. This can be written as ${}^4{C_3} \cdot {}^3{C_2} \cdot {}^2{C_1}$ .
The third way of choosing 6 questions can be 3 from section 1, 1 from section 2 and 2 from section 3. This can be written as ${}^4{C_3} \cdot {}^3{C_1} \cdot {}^2{C_2}$ .
The fourth way of choosing 6 questions can be 2 from section 1, 3 from section 2 and 3 from section 3. This can be written as ${}^4{C_2} \cdot {}^3{C_3} \cdot {}^2{C_1}$ .
The fifth way of choosing 6 questions can be 2 from section 1, 2 from section 2 and 2 from section 3. This can be written as ${}^4{C_2} \cdot {}^3{C_2} \cdot {}^2{C_2}$ .
The sixth way of choosing 6 questions can be 1 from section 1, 3 from section 2 and 2 from section 3. This can be written as ${}^4{C_1} \cdot {}^3{C_3} \cdot {}^2{C_2}$ .
All the above-mentioned combinations are independent from each other. So, to get the total number of ways we will add all the above combinations.
$= \left( {{}^4{C_4} \cdot {}^3{C_1} \cdot {}^2{C_1}} \right) + \left( {{}^4{C_3} \cdot {}^3{C_2} \cdot {}^2{C_1}} \right) + \left( {{}^4{C_3} \cdot {}^3{C_1} \cdot {}^2{C_2}} \right) + \left( {{}^4{C_2} \cdot {}^3{C_3} \cdot {}^2{C_1}} \right) + \left( {{}^4{C_2} \cdot {}^3{C_2} \cdot {}^2{C_2}} \right) + \left( {{}^4{C_1} \cdot {}^3{C_3} \cdot {}^2{C_2}} \right)$

$$= \left( {\dfrac{{4!}}{{4! \times 0!}} \times \dfrac{{3!}}{{1! \times 2!}} \times \dfrac{{2!}}{{1! \times 1!}}} \right) + \left( {\dfrac{{4!}}{{3! \times 1!}} \times \dfrac{{3!}}{{2! \times 1!}} \times \dfrac{{2!}}{{1! \times 1!}}} \right) + \left( {\dfrac{{4!}}{{3! \times 1!}} \times \dfrac{{3!}}{{1! \times 2!}} \times \dfrac{{2!}}{{2! \times 0!}}} \right) + \left( {\dfrac{{4!}}{{2! \times 2!}} \times \dfrac{{3!}}{{3! \times 0!}} \times \dfrac{{2!}}{{1! \times 1!}}} \right) + \\\ \left( {\dfrac{{4!}}{{2! \times 2!}} \times \dfrac{{3!}}{{2! \times 1!}} \times \dfrac{{2!}}{{2! \times 0!}}} \right) + \left( {\dfrac{{4!}}{{1! \times 3!}} \times \dfrac{{3!}}{{3! \times 0!}} \times \dfrac{{2!}}{{2! \times 0!}}} \right) \\\$$

$= \left( {1 \times 3 \times 2} \right) + \left( {4 \times 3 \times 2} \right) + \left( {4 \times 3 \times 1} \right) + \left( {6 \times 1 \times 2} \right) + \left( {6 \times 3 \times 1} \right) + \left( {4 \times 1 \times 1} \right)$
= 6 + 24 + 12 + 12 + 18 + 4
= 76
Thus, a candidate can choose 6 questions from 3 different sections, at least 1 from each, in 76 different ways.

Note: If the possible combinations or permutations are independent from each other, then by the additive property, we can add all of the possible permutations or combinations.
If the possible combinations or permutations depend on each other, then by the multiplicative property, we can multiply all of the possible permutations or combinations.