#what formula can i use for this question

31 messages · Page 1 of 1 (latest)

vital wagon
#

help!!

Capture.PNG
halcyon flowerBOT
#
  1. Wait patiently for a helper to come along.
  2. Once someone helps you, say thank you and close the thread with:
+close
  1. Feel free to nominate the person for helper of the week in #helper-nominations
  2. Do not ping the mods, unless someone is breaking the rules.
  3. If you're happy with the help you got here, and the server overall, you can contribute financially as well:
safe coral
#

oh, its just $n^m$ for there are n choices and m digit number you makee

junior pastureBOT
#

ultra racism

563022261809053696.png
safe coral
#

@vital wagon

cobalt dock
#

u can use permutation to solve this, make three boxes where u can fill 4 digits- 1, 2, 5, 6 and 8. now in the 1st box u can fill 4 digits and in the 2nd one, 4 and so on, therefore u can in total make 4^3 = 64 numbers (if repetition of numbers is allowed)

hollow plover
#

Repetition isnt allowed cause "by choosing three different numbers from the following list"

#

You have to think about "how many possibilities do you have for the first digit of your 3-digit numbers ?"

#

Then "How many possibilities for the second digit and then for the last digit"

plucky saffron
#

PnC that is permutation and combination

#

543

vital wagon
quaint gladeBOT
#

@vital wagon has given 1 rep to @safe coral

vital wagon
#

that is strange

safe coral
#

then use permutation

#

^

#

apparently it works for some cases

#

sry

#

didnt realize

#

🤡🤡

vital wagon
#

dw

#

im gonna solve it with my passion for writing shit code

#

im take 3 random numbes and rearrange them randomly if the comonation is unique then n += 1 if its not re do it and if this happens lets say 10,000 times without adding 1 to n then lets output the final awnser

wooden hemlock
wooden hemlock
vital wagon
#

im still gonna code

quaint gladeBOT
#

@vital wagon has given 1 rep to @wooden hemlock

dull sable
#

True should be 5 x 4 x 3