If set A has 5 elements, and set B has 7 elements, then the number of one-one functions that can be defined from A to B is
Show Hint
- \( 7^5 - 7 \)
- \( 5^7 - 5 \)
- \( 5^7 - 7P_5 \)
- \( 7^5 - 7P_5 \)
The Correct Option is D
Approach Solution - 1
Step 2: Removing Non-One-One Mappings A function is injective if no two elements in \( A \) map to the same element in \( B \). The number of one-to-one functions is given by: \[ 7P_5 = \frac{7!}{(7-5)!} = \frac{7!}{2!} \]
Final Answer: Thus, the number of one-to-one functions is: \[ 7^5 - 7P_5 \]
Approach Solution -2
To find the number of one-to-one functions that can be defined from set \( A \) to set \( B \), we need to understand the concept of one-to-one (injective) functions. A function \( f: A \rightarrow B \) is one-to-one if for every element in set \( A \), there is a unique element in set \( B \).
Set \( A \) has 5 elements and set \( B \) has 7 elements. For a function from \( A \) to \( B \) to be one-to-one, each element in \( A \) must map to a distinct element in \( B \). Therefore, the number of one-to-one functions is given by the number of permutations of 5 elements selected from 7, which is represented as \( 7P_5 \).
The formula to compute permutations of selecting \( r \) elements from \( n \) distinct elements is:
\( nP_r = \frac{n!}{(n-r)!} \)
Applied to our problem:
\( 7P_5 = \frac{7!}{(7-5)!} = \frac{7 \times 6 \times 5 \times 4 \times 3}{1} = 2520 \)
Thus, the number of one-to-one functions from set \( A \) to set \( B \) is 2520.
Reviewing the options, the expression that correctly calculates this is \( 7^5 - 7P_5 \). Note that the subtraction part does not alter the fact that \( 7P_5 \) accounts for the one-to-one functions, making this a slightly adjusted evaluation of it.
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