#help pls (permutation and combination)
1 messages · Page 1 of 1 (latest)
- Do not ping the Moderators, unless someone is breaking the rules.
- Do not ping the Helper Moderators, unless there is a conflict between helpers.
- Do not ping other members randomly for help.
- Ask your question and show the work you've done so far. If you've posted a screenshot of a question, specify which part you need help with.
- Wait patiently for a helper to come along.
- If the Helper has answered your question, remember to thank them with the Mathematics Ranks bot and close the thread with:
+close
Feel free to nominate the person for helper of the week in #helper-nominations
If you're happy with the help you got here, and the server overall, you can contribute financially as well:
well, let's start with part a.
we know that we can treat the smart watches as one block, and that the tablets shouldn't be next to each other.
then, do complimentary counting with the tablets, i.e. you count the total amount of arrangements where the tablets are together and subtract it from the total arrangements possible.
then just multiply the latter by the amount of permutations for the smart watches
if you need me to explain the intuition, you can ping me
wel, you can’t just give me some calculation without intuition
why is that the case?
t1 p1 t2 p2 t3 p3 t4 p4 t5 p5 t6 p6 w1 w2 w3 t7
7 option out of 3 must fill in tablet
?
@analog sigil
t means tablet p means phone and w means watches
Watches must to together so w1 w2 w3 together
yes
finding arrangements with tablets next to each other is going to be a bit taxing with 3 tablets. it would be easier to imagine first arranging the watches (as 1 block of 3) and phones, then inserting the tablets between them. you can imagine empty slots between them (and on the outside), and then you want to permute the tablets among these slots
but put them into one W block
Oh
like T1 T2 T3 S1...S5 and W
we want them to be together
t1 p1 t2 p2 t3 p3 t4 p4 t5 p5 t6 p6 w1-3 t7
no need o write it out
more than one possibility for them to not be next to each other
my point was to use complimentary counting
What is that
to count the permutations where tablets are together
then subtract the total amount of permutation from it
How to count?
put them together
11! / 3! x 7!
force them into a pair or triplet
?
one second
why?
11 items
Divide by 3 identical item (watches)
Then
Wait
Sorry
I'm confused by my own stuff😅
...
refer back to this, please
read the whole thing carefully
||for the phones and watches, you can permute them 3!*6! ways. then there are 7p3 ways you can insert the tablets between them (keeping the 3 watches together)||
basically you permute the phones and watches, p1, p2, p3, p4, p5 and w1w2w3. so that's 6 elements. but also you can permute w1, w2, w3. and then you permute the tablets amongst the empty spaces because you can have at most one tablet in each one
@waxen panther
Hello kizaru_6208, this is a friendly reminder that your help request has been inactive for more than 24 hours. If you no longer need assistance, please consider closing the thread using the +close command. This thread will be automatically closed in 3 days if it remains inactive.