How many 3-digit numbers can be formed sing the digits 0,1, 2, 3, 4, 5, 6, if repetitions are not allowed?
however 180 can't be the answer i checked? any one know how to solve this one
permutations = 7!/4! = 7*6*5 = 210
but does beginning with 0 count like 012 ?
210 was my choice answer and I got it wrong
10 in the ones column
* 9 in the tens column
* 7 in the hundreds column if 0 not allowed
90*7 = 630 ?
To find the number of 3-digit numbers that can be formed using the given digits without repetition, you can use the concept of permutations.
Permutations represent the number of ways to arrange a set of objects in a specific order. In this case, you have a set of 8 digits (0, 1, 2, 3, 4, 5, 6) and you need to find the number of ways you can arrange them to form a 3-digit number.
To solve this, you can use the formula for permutations:
P(n, r) = n! / (n - r)!
Where n is the total number of objects in the set, and r is the number of objects you want to choose.
In this case, you have 8 digits (n = 8) and you want to form a 3-digit number (r = 3).
Using the formula:
P(8, 3) = 8! / (8 - 3)!
= 8! / 5!
= (8 x 7 x 6 x 5!) / 5!
= 8 x 7 x 6
= 336
Therefore, there are 336 different 3-digit numbers that can be formed using the digits 0, 1, 2, 3, 4, 5, 6 without repetition.
If you checked and got a different answer, please double-check your calculations or provide more details about how you arrived at the answer of 180, and I would be happy to assist you further.