IF NO DIGIT MAY BE USED MORE THAN ONCE, HOW MANY 2-DIGIT NUMBERS CAN BE FORMED USING ONLY THE DIGITS 9, 6, 5, 7, 4, AND 2?
There are six choices for the first digit (any of the six given digits can be used), and five choices for the second digit (only five digits are left after choosing the first). Therefore, there are $6\times5=\boxed{30}$ 2-digit numbers that can be formed using only the digits $9, 6, 5, 7, 4,$ and $2$, with no digit used more than once.
so how many?
There are $\boxed{30}$ possible 2-digit numbers that can be formed using only the digits 9, 6, 5, 7, 4, and 2, with no digit used more than once.
To find the number of 2-digit numbers that can be formed using the digits 9, 6, 5, 7, 4, and 2 without repeating any digit, we can use the concept of permutations.
Permutation is a concept in combinatorics that represents the number of ways to arrange a set of objects in a specific order.
In this case, we need to find the number of permutations of 6 digits taken 2 at a time, without repetition.
The formula to calculate the number of permutations is:
P(n, r) = n! / (n - r)!
Where P(n, r) represents the number of permutations of n objects taken r at a time, and "!" denotes factorial.
Applying this formula to our problem, we have:
P(6, 2) = 6! / (6 - 2)!
= 6! / 4!
= (6 x 5 x 4 x 3 x 2 x 1) / (4 x 3 x 2 x 1)
= (720) / (24)
= 30
Therefore, there are 30 different 2-digit numbers that can be formed using the digits 9, 6, 5, 7, 4, and 2, without repeating any digit.