is there a better way to do this other than putting in indivisual values of n1 n2 n3 n4 n5 and tediously finding all 💀
#PnC Hit/Trial Alternative
lucid moon
halcyon wind
lucid moon
DUDE I WAS LEGIT ABoUT To PING YOUU
weak salmon
DUDE I WAS LEGIT ABoUT To PING YOUU
I legit just got out of English exam so don't expect much lol
lucid moon
oh damn
weak salmon
Ok, first just find number of integral solutions using the formula. Then, exclude all solutions where a number is repeated because strict inequality, then, divide by five because we need only the ascending order solutions.
For repeated solutions, the largest one will be 4,4,4,4,4
So you just need to check how many permutations you have with 2 of 1,2,3,4, with 3 of 1,2,3,4 and with 4 of 1,2,3
Tedious but not impossible
That's my method. @spice scroll might have a better one.
lucid moon
Ok, first just find number of integral solutions using the formula. Then, exclud...
- what formula are you talking about 💀
- DIVIDING BY 5 DUDE THAT IS SO SMART IT DIDNT EVEN CROSS ME THAT IT COULD BE DONE THAT WAY
lucid moon
For repeated solutions, the largest one will be 4,4,4,4,4
So you just need to ch...
this didnt make much sense tbh
weak salmon
1. what formula are you talking about 💀
2. DIVIDING BY 5 DUDE THAT IS SO SMART...
$\binom{n-r+1}{r+1}$
hollow heraldBOT
Opt
weak salmon
Oops
lucid moon
that was for multiples as far as i remember one sec
weak salmon
Grouping ka formula.
lucid moon
yea i dont think i know that where can i learn that
junior gate
is there a better way to do this other than putting in indivisual values of n1 n...
you just need to find no of unique solutions when n1 n2 ..n5 are not equal and each set of n1 n2 ..n5 will have only 1 way to arrange them.
hard cairn
strict inequality nikalne ke liye we used to add 1 to each number, if i remember right
weak salmon
strict inequality nikalne ke liye we used to add 1 to each number, if i remember...
No that's for non-negative and positive solutions
hard cairn
ohh ok
lucid moon
ok so isko apply kaise karenge idhar
lucid moon
you just need to find no of unique solutions when n1 n2 ..n5 are not equal and ...
how-
junior gate
how-
let's take a simpler problem n1+n2+n3 =7 (n1<n2<n3) 1 2 4 is a solution and there is only one way to arrange it such that n1<n2<n3
lucid moon
ok goteem
spice scroll
That's my method. <@1075951732460376214> might have a better one.
Unless you know partitions, counting is the only way ig.
lucid moon
Unable to make out more than these I need 2 more
