How many different 3-digit numbers can you make with the following digits if no digit may be repeated in the same number?
2 7 8 0 3 1
Is there any way I can solve this problem by writing equation?
there are 5 choices for the 1st (leftmost) digit
... a "leading" zero is not standard form
5 choices for the 2nd digit
... zero is available
4 choices for the 3rd digit
5 * 5 * 4 = ?
Answer is 6*5*4=120
Thank you anyway!
Yes, you can solve this problem using a simple equation. To find the number of different 3-digit numbers you can make without repeating any digit, you can use the concept of permutations.
The number of permutations of n objects taken r at a time is given by the equation:
P(n, r) = n! / (n - r)!
Where n! represents the factorial of n, which is the product of all positive integers from 1 to n.
In this case, we have 6 digits to choose from (2, 7, 8, 0, 3, and 1) and we need to choose 3 digits to form a 3-digit number.
Therefore, using the equation P(6, 3), we can solve for the number of different 3-digit numbers as follows:
P(6, 3) = 6! / (6 - 3)!
= 6! / 3!
= (6 x 5 x 4) / (3 x 2 x 1)
= 120 / 6
= 20
So, there are 20 different 3-digit numbers that can be formed using the given digits without repeating any digit.