A combination lock has 25 numbers on it. How many different 3-digit lock combinations are possible if no digit can be repeated?

Mathematics

1 answer:

IgorC [24]3 years ago

7 0

Answer:

The number of ways is 13,800 ways

Step-by-step explanation:

In this question, we are tasked with calculating the number of ways to form 3-digits codes from 25 different numbers given that no repetition is allowed.

Now, for the first digit, we have all the 25 numbers completely wanting to take a spot. Now, in how many ways can we choose a single out of 25, that would be 25 ways

Now, we have 24 numbers left, and we are trying to pick one, the number of ways this can be done is 24 ways also

for the last digit, we have 23 numbers and we are selecting just one, the number of ways this can be done is 23 ways too

Thus, the cumulative number of ways would be 25 * 24 * 23 = 13,800 ways

You might be interested in

Write the equation of an ellipse with the following characteristics: center =(1,-3) vertices=(-6,-3),(8,-3)

leonid [27]

Answer:

(x-1)²/49 + (y+3)²/16.51 = 1

Step-by-step explanation:

vertex (8,-3) Foci (6.7,-3) center (h,k) = (1,-3) , major axis parallel to x axis

a is the distance of the vertex from the center, and c is the distance of the foci from the center. b is semi-minor axis

b² = a² - c² Ellipse equation: (x-h)²/a² + (y-k)²/b² = 1

a = 8-1 = 7 c = 6.7 - 1 = 5.7

b² = 7² - 5.7² = 16.51

Ellipse equation: (x-1)²/49 + (y- -3)²/16.51 = 1 i.e. (x-1)²/49 + (y+3)²/16.51 = 1

5 0

3 years ago

Read 2 more answers

Find the length of the missing side

faltersainse [42]

Answer:

c = 17

Step-by-step explanation:

Since this is a right angle triangle we can use the Pythagoras theorem that states that $c^{2} = a^{2} + b^{2}$ (where c is the Solve for hypotenuse and "a" and "b" are the legs of the right angle triangle). From here we know that the answer to this question is....