If no digit may be used more than once, how many 2-digit numbers can be formed using only the digits 5, 6, 3, and 2?
11
12
10
2
To find the number of 2-digit numbers that can be formed using the digits 5, 6, 3, and 2 without repeating any digit, we need to count the number of choices for each digit.
The first digit can be chosen from any of the four available digits (5, 6, 3, and 2). So there are 4 choices for the first digit.
The second digit can be chosen from the remaining three available digits (since we cannot repeat any digit). So there are 3 choices for the second digit.
Therefore, the total number of 2-digit numbers that can be formed is obtained by multiplying the number of choices for each digit: 4 choices for the first digit multiplied by 3 choices for the second digit.
4 choices for the first digit * 3 choices for the second digit = 12
So, there are 12 different 2-digit numbers that can be formed.
To determine the number of 2-digit numbers that can be formed using the digits 5, 6, 3, and 2 without repetition, we need to use the concept of permutation.
In this case, we have 4 available digits (5, 6, 3, and 2) and we need to choose 2 digits to form a 2-digit number.
The formula for the number of permutations without repetition is given by: P(n, r) = n! / (n - r)!
Where:
- P(n, r) represents the number of permutations of r objects chosen from a set of n objects.
- n! denotes the factorial of n, which is the product of all positive integers from 1 to n.
Applying the formula, we can calculate the number of 2-digit numbers that can be formed:
P(4, 2) = 4! / (4 - 2)!
= 4! / 2!
= (4 x 3 x 2 x 1) / (2 x 1)
= 24 / 2
= 12
Therefore, there are 12 different 2-digit numbers that can be formed using the digits 5, 6, 3, and 2 without repetition.
So the correct answer is 12.
56
53
52
65
35
25
63
62
36
26
23
32
Did I list all of the possibilities?
There are 4 digits to fill 2 spaces
number = (4)(3) = 12