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Step-by-Step Solution
Step 1: Interpret the Problem
We have two sets:
R = \{a, b, c, d, e\} (5 elements) and S = \{1, 2, 3, 4\} (4 elements).
We want to count the total number of onto (surjective) functions
f : R \to S such that f(a) \neq 1 .
Step 2: Count All Onto Functions from R to S
An onto function from a 5-element set to a 4-element set means each of
the 4 elements in S appears as an image at least once. The number of all
functions from R to S is 4^5 = 1024 . However, to count only onto
functions, we use the principle of inclusion-exclusion:
\text{Total onto}
= 4^5
- \binom{4}{1} 3^5
+ \binom{4}{2} 2^5
- \binom{4}{3} 1^5.
Explicitly,
4^5 = 1024, \quad
3^5 = 243, \quad
2^5 = 32, \quad
1^5 = 1.
Then,
\text{Total onto}
= 1024
- \binom{4}{1} \cdot 243
+ \binom{4}{2} \cdot 32
- \binom{4}{3} \cdot 1
= 1024
- 4 \times 243
+ 6 \times 32
- 4 \times 1
= 240.
So there are 240 onto functions in total from R to S .
Step 3: Count Onto Functions Where f(a) = 1
Now we must count how many onto functions from R to S have the specific
condition f(a) = 1 . In this scenario, for the function to still be onto,
the remaining four elements \{b, c, d, e\} must collectively map onto
\{1, 2, 3, 4\} in such a way that 2, 3, and 4 each appear at least once
among the images of \{b, c, d, e\} .
The total number of ways to map \{b, c, d, e\} to \{1, 2, 3, 4\} is 4^4 = 256.
We exclude mappings that fail to include at least one of 2, 3, 4 .
Using inclusion-exclusion again:
Subtract the number of mappings that miss at least one of 2, 3, or 4 :
\binom{3}{1} \times 3^4 = 3 \times 81 = 243.
Add the number of mappings that miss at least two of them (say, missing 2 and 3 , or 2 and 4 , or 3 and 4 ):
\binom{3}{2} \times 2^4 = 3 \times 16 = 48.
Subtract the case that misses all three of them (i.e., uses only 1 ):
\binom{3}{3} \times 1^4 = 1.
Hence, the number of ways to ensure that 2, 3, and 4 appear at least once is:
256 - 243 + 48 - 1 = 60.
Step 4: Apply the Restriction f(a) ≠ 1
We found that among the 240 onto functions, there are 60 onto functions
with f(a) = 1 . Therefore, the number of onto functions satisfying
f(a) \neq 1 is:
240 - 60 = 180.
Final Answer
The total number of onto functions f : R \to S such that
f(a) \neq 1 is \boxed{180} .
