Greetings!
When attempting to solve this problem I originally got 6!/3!3! as the number of possible combinations… just like in the answer key. I set it up like
555XXX which had 3 repeats and 3 repeats yielding 3!3! in the denominator.
But then I second guessed myself and said that the X values could be any number on the die besides 5, which wouldn’t necessarily be repeating.
Because of this I eliminated the second 3! in the denominator and got 6!/3! which is 120.
What is the logic of using the second 3! in the denominator as indicated in the answer key?
Thank you as always for your assistance everyone!
Similar question on the following quiz question #2.
When I setup the combination which I’m basing the Total!/Repeat! I get
335XX from which I got
5!/2!.. vs answer key 5!/3!2!
What am I missing here?
Hi leaderboard! Thanks for checking in - the questions are:
That is correct. It also doesn’t matter what the Xs are, as they aren’t really relevant (except that they are not 5s). However: the 5s themselves are repeating, which is why we need to divide by 3!. Notice that this is permutations vs combinations, as we don’t really care on what the order of the 5s themselves are.
Hi Leaderboard!
Thanks for your reply mate!
Got it ok… but I’m still confused why we would divide 6!/3!3! instead of just 6!/3!
The 5’s are repeating which counts as 1x 3! … where does the second 3! come from since the XXX could be ANY set of numbers other than 5?
Preciate your help as always
Recall that
\frac{n!}{r!}
is permutations
while
\binom{n}{r} = \frac{n!}{r! \times (n - r)!}
is combinations - do you now see why?
Ahah - got it thank you! Need to crank out some Quant Flashcards to start drilling these
Thanks for your assistance!
