A lock has a code of 3 numbers between 1 and 15. If no numbers in the code are allowed to repeat, how many different codes could be made?
15P3 = 2730
Oh, the possibilities are endless! Well, maybe not endless, but there are definitely quite a few. Let's see, if we have 3 numbers to choose from between 1 and 15, and none of them can repeat, then for the first number, we have 15 options. For the second number, we have 14 options left (since we can't repeat the first number). And for the third number, we have 13 options left. So, if we multiply these choices together, we get 15 * 14 * 13 = 2730! That's a lot of different codes you can make. Just make sure you don't forget which one you choose, or you might be stuck clowning around with a locked lock forever!
To find out the number of different codes that could be made with the given conditions, we need to use the concept of combinations.
In this case, we have to select 3 numbers between 1 and 15 without repeating any number.
To calculate this, we can use the formula for combinations. The formula for combinations is:
C(n, r) = n! / (r! * (n-r)!)
Where:
- n is the total number of options available (15 in our case)
- r is the number of options to be chosen (3 in our case)
- ! represents the factorial operation (the product of all positive integers from 1 to the given number)
Let's calculate it using the formula:
C(15, 3) = 15! / (3! * (15-3)!)
= 15! / (3! * 12!)
Using factorials, the calculation becomes:
C(15, 3) = (15 * 14 * 13 * 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1) / [(3 * 2 * 1) * (12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1)]
Now we can simplify this expression:
C(15, 3) = (15 * 14 * 13) / (3 * 2 * 1)
= 2,730 / 6
= 455
Therefore, there are 455 different codes that could be made with the given conditions.
To find the number of different codes that can be made, we can use the concept of combinations.
Since the lock has 3 numbers and no repetition is allowed, we need to select 3 numbers from a pool of 15 numbers.
The number of ways to select "r" objects from a set of "n" distinct objects is given by the formula for combinations:
C(n, r) = n! / (r! * (n-r)!)
In this case, n = 15 (the number of available numbers) and r = 3 (the number of numbers to be selected).
So, the number of different codes that can be made is:
C(15, 3) = 15! / (3! * (15-3)!)
= 15! / (3! * 12!)
= (15 * 14 * 13) / (3 * 2 * 1)
= 455
Therefore, there are 455 different codes that can be made for the lock.