Question

In an examination a candidate has to pass in each of four subjects. In how many ways can he fail?
A) 15
B) 20
C) 25
D) 10

Answer 1

Here the given question is based on the concept of combination. We have to find the number of ways a candidate has to fail. For this, first we need a number of possible ways to fail in each subject by using the formula of combination and further by adding all the ways to get the required solution.

Complete step by step answer:
Combination is defined as “the arrangement of ways to represent a group or number of objects by selecting them in a set and forming the subsets”
Generally, combination denoted by ‘$^n{C_r}$’, \left( {\begin{array}{*{20}{c}} n \\\ r \end{array}} \right), or ‘$n$ choose $r$’
The formula used to calculate the combination is: $^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$----(1)
Where, ‘$n$’ is the number of items you have to choose from (total number of objects) and ‘$r$’ is the number of items you're going to select.
Now consider the question,
In a examination a candidate has to pass in each of four subjects i.e., $n = 4$
We need to find the number of ways the student can fail.
A candidate can be failed by failing in one paper or in two papers or in three papers or in all the four papers.
If a candidate can be fails in one paper, the number of ways = $^4{C_1}$
If a candidate can be fails in two papers, the number of ways = $^4{C_2}$
If a candidate can be fails in three papers, the number of ways = $^4{C_3}$
If a candidate can be fails in all four papers, the number of ways = $^4{C_4}$
The total number of ways can candidate to be fails is:
$\Rightarrow \,\,{\,^4}{C_1} + \,{\,^4}{C_2} + \,{\,^4}{C_3} + + \,{\,^4}{C_4}$
Now apply a formula of combination, then
$\Rightarrow \,\,\,\dfrac{{4!}}{{\left( {4 - 1} \right)! \cdot 1!}} + \,\,\dfrac{{4!}}{{\left( {4 - 2} \right)! \cdot 2!}} + \,\,\dfrac{{4!}}{{\left( {4 - 3} \right)! \cdot 3!}} + \dfrac{{4!}}{{\left( {4 - 4} \right)! \cdot 4!}}$
$\Rightarrow \,\,\,\dfrac{{4 \times 3!}}{{3! \cdot 1!}} + \,\,\dfrac{{4 \times 3 \times 2!}}{{2! \cdot 2!}} + \,\,\dfrac{{4 \times 3!}}{{1! \cdot 3!}} + \dfrac{{4!}}{{0! \cdot 4!}}$
$\Rightarrow \,\,\,\dfrac{{4 \times 3!}}{{3! \cdot 1}} + \,\,\dfrac{{4 \times 3 \times 2!}}{{2! \cdot \left( {2 \times 1} \right)}} + \,\,\dfrac{{4 \times 3!}}{{1 \cdot 3!}} + \dfrac{{4!}}{{1 \cdot 4!}}$
On simplification, we get
$\Rightarrow \,\,\,4 + \,\,\dfrac{{12}}{2} + \,\,4 + 1$
$\Rightarrow \,\,\,4 + \,\,6 + \,\,4 + 1$
$\therefore \,\,\,15\,ways$
Hence, the number of ways that the student can fail is $15\,ways$. Therefore, option (A) is the correct answer.

Note:
In combinations each of the different selections made by taking some or all of a number of objects irrespective of their arrangement. Remember, factorial is the continued product of first n natural numbers is called the “n factorial “ and it represented by$n! = \left( {n - 1} \right) \cdot \left( {n - 2} \right) \cdot \left( {n - 3} \right).....3 \cdot 2 \cdot 1$ and value of $0! = 1$.