Combinations Without Repetition Example
Combination without repetition example of the conference shows the formula for a Variation as V(10 3) and then proceeds to solve this by doing 8*9*10 = 720. Accoring to the previous lesson on Variations with repetition, the V(10 3) solution should be 4*5*6*7*8*9*10.
It was originally taught as using the numbers that are different between the Options and the Positions in the factorial, so in this case 4-10. But, the lesson noted above is using the last 3 number of the Options count instead, or 8-10.
Which is correct?
It was originally taught as using the numbers that are different between the Options and the Positions in the factorial, so in this case 4-10. But, the lesson noted above is using the last 3 number of the Options count instead, or 8-10.
Which is correct?
1 answers ( 0 marked as helpful)
Seems like I just needed to rubber duck this. I was missing a step in the formula solving process for a Variation without Repetition. Specifically, I wasn't doing the (n-p)! subraction correctly. I should have done 10!(10-3)! to get 10!/7!. Once I did that step correctly, I ended up with the remaining numbers to factorialize as 8,9 and 10.
I would definitely be up for an exponential increase in cost if there were times available to ask questions of an instructor on a daily or semi-daily basis. Every class seems to have something like this that could save hours of confusion, or in some cases just a flat-out lack of understanding that I have to accept and move on.
Thanks.
I would definitely be up for an exponential increase in cost if there were times available to ask questions of an instructor on a daily or semi-daily basis. Every class seems to have something like this that could save hours of confusion, or in some cases just a flat-out lack of understanding that I have to accept and move on.
Thanks.