how many combonations?

Question

for the numbers 2 4 5 6 7 8 0

four digit combonation

cant use a number twice

zero at the end

Answer 1

The answer is 120 "lock combos" if I've understood your question.

You can't use a number twice, but different orders are allowed, but a combo must end in 0. So, for example, 2450 and 5240 are valid combo numbers. If that's the case, the answer is 120.

You must choose 3 distinct numbers in any order from the numbers 2 4 5 6 7 8. You then add a 0 on the end of these 3 to get a valid 4 digit combo number. The number of ways of doing this is simply 6x5x4 = 120. Note, if digits could be used more than once it would 6x6x6 = 216.
As far as using formulas go, you would use the permutation formula:

nPr = n!/(n-r)! In your case n= 6 and r = 3

6P3 = 6x5x4 = 120

Answer 2

7! x 6! x 5! x 4!


7! (7 x 6 x 5 x 4 x 3 x 2 x 1)

Answer 3

since you specified 0 at the end, the problem simplifies to
3 spots for 6 digits.

6*5*4*1
120.

Answer 4

3^6

Answer 5

possible for last digit: 1 (only 0)
possible for first digit: 6
possible for 2nd digit: 5
possible for 3rd digit: 4

possible 4-digit combination where last digit is zero:
1 x 6 x 5 x 4 = 120

^_^

Answer 6

nCr = (n!)/(k!(n - k)!)

7C4 = (7!)/(4!(7 - 4)!)
7C4 = (5040)/(24 * 3!)
7C4 = 5040/(24 * 6)
7C4 = 5040/144
7C4 = 35

Assuming 0234 is allowed

Answer 7

2450 to many
4560
5670
6780