pNc
wont ans be 6C4 x 4! x 4x3
first select 4 ppl , then arrange by 4! then the 2 ppl left have 4 and 3 choices
ans is 1080 by mine is coming out to be 4320
11 Replies
@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.The only case here is people distributed as 1,2,2,1
So, 4c2.6c2.4c2.2c1 would be the required value ig.
(Assuming the rooms are numbered)
The overcounting occurs because for instance let P1 "enter one of the four rooms by 6c4" and P3 "enter the same room by x4 term" then the reverse (P3 by 6c4 and P1 by x4 factor) is also counted.
The amount of overcounting because of the x4 term would be, hence, by a factor of 2 and by x3 by a factor of 2.
Hence the overcounting gives a factor of 4 in this case.
oh shit
what about 4!/2!x2! for arrangemen
multiply by 6 for arrangement for rooms?
That is the first term in this expression
For arrangements of 1,2,2,1 distribution
wait so first u selected 2 ppl , then again 2 ppl , then 1 person , and the last person gets the room left by default?
Yep. Or you could use the distribution formula.
whats that?
that divide m , into n groups?
Yep.
ic
thx
+solved @SirLancelotDuLac
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