a lock has a code of 5 numbers between 1 and 20 . If no numbers in the code are allowed to repeat, how many different codes could be made?
so is the answer
of 20C5
then 15,504?
no ... the earlier posting was corrected
Scott, please just explain in baby words..(give me this answer) and I can figure out the other 9 questions.
This blended school learning is not for me....
thanks a million
my parents have no clue either
really stuck
i got it!
Yes, you are correct. To find the number of different codes that could be made, you need to calculate the number of combinations of choosing 5 numbers out of 20 without repetition. This can be expressed as "20 choose 5" or written as 20C5.
The formula for combinations is given by:
nCk = n! / (k! * (n-k)!)
Where nCk represents the number of combinations of choosing k items from a set of n items without repetition, and ! denotes factorial (the product of all positive integers up to and including the number).
In this case, you want to find 20C5. Plugging the values into the formula, we get:
20C5 = 20! / (5! * (20-5)!)
Calculating the factorials:
20! = 20 * 19 * 18 * 17 * 16 * 15 * 14 * 13 * 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
5! = 5 * 4 * 3 * 2 * 1
(20-5)! = 15! = 15 * 14 * 13 * 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
Simplifying the equation:
20C5 = 20! / (5! * 15!)
Now, calculating the factorials and simplifying the equation further:
20C5 = (20 * 19 * 18 * 17 * 16 * 15!) / (5! * 15!)
The 15! terms cancel out:
20C5 = (20 * 19 * 18 * 17 * 16) / (5!)
Finally, calculating 5!:
5! = 5 * 4 * 3 * 2 * 1 = 120
Substituting this value back into the equation:
20C5 = (20 * 19 * 18 * 17 * 16) / 120
After performing the calculation:
20C5 = 15,504
So, the number of different codes that could be made is indeed 15,504.