In an examination a candidate has to pass in each of the 5 subjects. In how many ways he can fail ?

Concept:

The number of ways to select r things out of n given things wherein r ≤ n is given by: \({\;^n}{C_r} = \frac{{n!}}{{r!\; \times \left( {n - r} \right)!}}\)

Calculation:

Given: There are 5 subjects and for a student to pass in an examination the students has to pass in each of the 5 subjects.

Here, we have to find in how many ways a student can fail in the examination.

In order to fail in the examination the student can fail in 1 or 2 or 3 or 4 or 5 subjects out of the 5 subjects in each case.

Case 1: Student fails in any 1 subject out of the 5 subjects.

No. of ways in which student can fail in any 1 subject out of the 5 subjects = \({\;^5}{C_1}\)

Case 2: Student fails in any 2 subjects out of the 5 subjects.

No. of ways in which student can fail in any 2 subjects out of the 5 subjects = \({\;^5}{C_2}\)

Case 3: Student fails in any 3 subjects out of the 5 subjects.

No. of ways in which student can fail in any 3 subjects out of the 5 subjects = \({\;^5}{C_3}\)

Case 4: Student fails in any 4 subjects out of the 5 subjects.

No. of ways in which student can fail in any 4 subjects out of the 5 subjects = \({\;^5}{C_4}\)

Case 5: Student fails in all 5 subjects.

No. of ways in which student can fail all 5 subjects. = \({\;^5}{C_5}\)

∴ Total number of ways in which a student can fail in an examination = \({\;^5}{C_1}\) + \({\;^5}{C_2}\) + \({\;^5}{C_3}\) + \({\;^5}{C_4}\) + \({\;^5}{C_5}\) = 31