The total number of injections (one-one and into mappings) form `{a_(1),a_(2),a_(3),a_(4)}` to `{b_(1),b_(2),b_(3),b_(4),b_(5),b_(6),b_(7)}` is

Options

A. 400
B. 420
C. 800
D. 840

Explanation

To find the total number of injections (one-one and into mappings) from the set \( A = \{a_1, a_2, a_3, a_4\} \) to the set \( B = \{b_1, b_2, b_3, b_4, b_5, b_6, b_7\} \), we can use the formula for permutations.

### Step-by-Step Solution:

1. **Identify the Sets**:
- Set \( A \) has 4 elements: \( |A| = 4 \).
- Set \( B \) has 7 elements: \( |B| = 7 \).

2. **Understand the Concept of Injections**:
- An injection (or one-one mapping) from set \( A \) to set \( B \) means that each element of \( A \) must map to a unique element in \( B \), and no two elements in \( A \) can map to the same element in \( B \).

3. **Use the Permutation Formula**:
- The number of injections from a set with \( R \) elements to a set with \( N \) elements is given by the formula \( N P R \), which is the number of ways to choose \( R \) elements from \( N \) and arrange them.
- Here, \( N = 7 \) (elements in set \( B \)) and \( R = 4 \) (elements in set \( A \)).
- The formula for permutations is:
\[ N P R = \frac{N!}{(N-R)!} \]

4. **Calculate \( 7 P 4 \)**:
- Using the formula:
\[ 7 P 4 = \frac{7!}{(7-4)!} = \frac{7!}{3!} \]
- Calculate \( 7! \) and \( 3! \):
\[ 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \]
\[ 3! = 3 \times 2 \times 1 = 6 \]
- Now substitute back into the permutation formula:
\[ 7 P 4 = \frac{5040}{6} = 840 \]

5. **Conclusion**:
- The total number of injections from set \( A \) to set \( B \) is \( 840 \).

### Final Answer:
The total number of injections from \( \{a_1, a_2, a_3, a_4\} \) to \( \{b_1, b_2, b_3, b_4, b_5, b_6, b_7\} \) is \( 840 \).

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