Question

Let \( R = \{a, b, c, d, e\} \) and \( S = \{1, 2, 3, 4\} \). Total number of onto functions \( f: R \to S \) such that \( f(a) \neq 1 \), is equal to:

• A. 240
• B. 180
• C. 204
• D. 216

Hint:

When counting onto functions with restrictions, calculate the total onto functions first and subtract the restricted cases using inclusion-exclusion principles.

Answer 1

The Correct Option is B

The correct answer is (B) : 180
Total no. of onto functions 
\(=\frac{5!}{3!2!}\times4!\)
So , when f(a) = 1
\(\frac{4!}{2!2!}\times3!+4!\)
\(\therefore\) Required functions : 
= 240 -36 -24
=180

Answer 2

The total number of onto functions from \( R \) to \( S \) is calculated as:

\[ \text{Total onto functions} = \binom{5}{3} \cdot 4! = \frac{5 \cdot 4}{2} \cdot 24 = 240. \]

Now, consider the case where \( f(a) = 1 \).

If \( f(a) = 1 \), the remaining 4 elements \( b, c, d, e \) must map onto \( S \setminus \{1\} \), which has 3 elements. The number of onto functions for these remaining 4 elements is:

\[ \text{Functions with \( f(a) = 1 \)} = \binom{4}{2} \cdot 3! \cdot 3. \]

Compute this step by step:

\[ \binom{4}{2} \cdot 3! = \frac{4 \cdot 3}{2} \cdot 6 +14= 60. \]

Finally, subtract this from the total:

\[ \text{Required functions} = 240 - 60 = 180. \]

Thus, the total number of onto functions \( f \) such that \( f(a) \neq 1 \) is \( \boxed{180} \).