A lock has a 3-number code made up of 20 numbers. If none of the numbers are allowed to repeat, how many different ways can you choose three different numbers in order for a unique code?
There are 20 choices for the first number, 19 choices for the second number (since one number has already been chosen), and 18 choices for the third number. Therefore, the number of ways to choose three different numbers from 20 without repetition is:
20 × 19 × 18 = 6840
So there are 6840 different ways to choose a unique 3-number code.
There are 20 numbers to choose from for the first position, 19 for the second position, and 18 for the third position since we can't repeat any numbers. So the total number of unique codes is 20 * 19 * 18 = 6840.
To find the number of different ways to choose three different numbers for a unique code, you can use the concept of combinations.
The formula to calculate combinations is nCr = n! / (r! * (n - r)!), where n is the total number of items and r is the number of items to be chosen.
In this case, there are 20 numbers to choose from, and we want to choose 3 numbers to form a unique code. Therefore, n = 20 and r = 3.
Applying the formula, we have:
nCr = 20! / (3! * (20 - 3)!)
Simplifying further:
20! = 20 * 19 * 18 * 17 * ... * 3 * 2 * 1
3! = 3 * 2 * 1
(20 - 3)! = 17 * 16 * ... * 3 * 2 * 1
nCr = (20 * 19 * 18 * 17 * 16 * 15 * 14 * 13 * 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1) / (3 * 2 * 1 * (17 * 16 * 15 * 14 * 13 * 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1))
Calculating the numerical value:
nCr = 1140
Therefore, there are a total of 1140 different ways to choose three different numbers for a unique code.
To find the number of different ways to choose three different numbers in order for a unique code, we can use the concept of combinations.
When selecting three numbers out of 20 without repetition, the order in which we choose the numbers doesn't matter. For example, selecting the numbers 5, 10, and 15 is the same as selecting the numbers 10, 5, and 15.
The formula to calculate the number of combinations is given by nCk = n! / (k! * (n - k)!), where n is the total number of elements and k is the number of elements to be chosen.
In this case, we need to calculate 20C3, which can be calculated as follows:
20! / (3! * (20 - 3)!)
= 20! / (3! * 17!)
To simplify the calculation, let's break down the factorials:
20! = 20 * 19 * 18 * 17!
3! = 3 * 2 * 1
Substituting these values back into the equation:
(20 * 19 * 18 * 17!) / (3 * 2 * 1 * 17!)
Simplifying further, we cancel out the common term "17!":
= (20 * 19 * 18) / (3 * 2 * 1)
Evaluating this expression gives us:
= (6840) / 6
= 1140
Therefore, there are 1140 different ways to choose three different numbers in order for a unique code on a lock with a 3-number code made up of 20 numbers where none of the numbers are allowed to repeat.