Question:
Let \( A = \{1, 3, 7, 9, 11\} \) and \( B = \{2, 4, 5, 7, 8, 10, 12\} \). Then the total number of one-one maps \( f: A \to B \), such that \( f(1) + f(3) = 14 \), is:
Let \( A = \{1, 3, 7, 9, 11\} \) and \( B = \{2, 4, 5, 7, 8, 10, 12\} \). Then the total number of one-one maps \( f: A \to B \), such that \( f(1) + f(3) = 14 \), is:
Updated On: Mar 25, 2026
- 180
- 120
- 480
- 240
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The Correct Option is D
Approach Solution - 1
Given sets:
\[ A = \{1, 3, 7, 9, 11\} \quad \text{and} \quad B = \{2, 4, 5, 7, 8, 10, 12\} \]
\[ A = \{1, 3, 7, 9, 11\} \]
\[ B = \{2, 4, 5, 7, 8, 10, 12\} \]
\[ f(1) + f(3) = 14 \]
(i) \( 2 + 12 \)
(ii) \( 4 + 10 \)
\[ 2 \times (2 \times 5 \times 4 \times 3) = 240 \]
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Approach Solution -2
Step 1: Understand the constraint on f(1) and f(3)
We require a one–one (injective) map f: A → B with A = {1, 3, 7, 9, 11}, B = {2, 4, 5, 7, 8, 10, 12}, such that f(1) + f(3) = 14.
List all unordered pairs from B that sum to 14:
2 + 12 = 14 (both in B) → valid;
4 + 10 = 14 (both in B) → valid;
5 + 9 = 14 (but 9 ∉ B) → invalid;
6 + 8 = 14 (but 6 ∉ B) → invalid;
7 + 7 = 14 (requires the same image for 1 and 3, which is impossible for a one–one map).
Therefore, the only usable unordered pairs are {2, 12} and {4, 10}.
Step 2: Count assignments for f(1) and f(3)
For each valid unordered pair, there are 2 ordered assignments (swap which value goes to 1 or 3):
Pair {2, 12}: (f(1), f(3)) = (2, 12) or (12, 2) → 2 ways;
Pair {4, 10}: (f(1), f(3)) = (4, 10) or (10, 4) → 2 ways.
Total ways to assign images of 1 and 3 = 2 + 2 = 4.
Step 3: Map the remaining elements injectively
After fixing f(1) and f(3), two elements of B are used. The remaining domain elements are {7, 9, 11} (3 elements), and the remaining codomain choices are 5 elements of B (7 total minus the 2 used).
Number of injective maps from a 3–element set into a 5–element set is the permutation P(5, 3) = 5 × 4 × 3 = 60.
Step 4: Multiply choices
For each of the 4 choices for (f(1), f(3)), there are 60 ways to assign f(7), f(9), f(11) injectively to the remaining 5 elements of B.
Hence, total valid one–one maps = 4 × 60 = 240.
Final answer
We require a one–one (injective) map f: A → B with A = {1, 3, 7, 9, 11}, B = {2, 4, 5, 7, 8, 10, 12}, such that f(1) + f(3) = 14.
List all unordered pairs from B that sum to 14:
2 + 12 = 14 (both in B) → valid;
4 + 10 = 14 (both in B) → valid;
5 + 9 = 14 (but 9 ∉ B) → invalid;
6 + 8 = 14 (but 6 ∉ B) → invalid;
7 + 7 = 14 (requires the same image for 1 and 3, which is impossible for a one–one map).
Therefore, the only usable unordered pairs are {2, 12} and {4, 10}.
Step 2: Count assignments for f(1) and f(3)
For each valid unordered pair, there are 2 ordered assignments (swap which value goes to 1 or 3):
Pair {2, 12}: (f(1), f(3)) = (2, 12) or (12, 2) → 2 ways;
Pair {4, 10}: (f(1), f(3)) = (4, 10) or (10, 4) → 2 ways.
Total ways to assign images of 1 and 3 = 2 + 2 = 4.
Step 3: Map the remaining elements injectively
After fixing f(1) and f(3), two elements of B are used. The remaining domain elements are {7, 9, 11} (3 elements), and the remaining codomain choices are 5 elements of B (7 total minus the 2 used).
Number of injective maps from a 3–element set into a 5–element set is the permutation P(5, 3) = 5 × 4 × 3 = 60.
Step 4: Multiply choices
For each of the 4 choices for (f(1), f(3)), there are 60 ways to assign f(7), f(9), f(11) injectively to the remaining 5 elements of B.
Hence, total valid one–one maps = 4 × 60 = 240.
Final answer
240
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