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.
If the domain of the function \[ f(x)=\log\left(10x^2-17x+7\right)\left(18x^2-11x+1\right) \] is $(-\infty,a)\cup(b,c)\cup(d,\infty)-\{e\}$, then $90(a+b+c+d+e)$ equals
Which of the following are ambident nucleophiles?
[A.] CN$^{\,-}$
[B.] CH$_{3}$COO$^{\,-}$
[C.] NO$_{2}^{\,-}$
[D.] CH$_{3}$O$^{\,-}$
[E.] NH$_{3}$
Identify the anomers from the following.

The standard Gibbs free energy change \( \Delta G^\circ \) of a cell reaction is \(-301 { kJ/mol}\). What is \( E^\circ \) in volts?
(Given: \( F = 96500 { C/mol}\), \( n = 2 \))