jitender lakhanpal
Last Activity: 12 Years ago
Dear Nikhil,
there can be three cases when n<r then the number of ways are 0 as there will be atleast one group that will be vacant
when n = r
then the number of ways are n factorial
when n>r then
the number of ways would be P(n,r) as there would be atleast one object in one group and then remaining n-r objects have to be placed in r things now there would be further 2 cases as
case:1 n-r>r then the total number of cases would be P(n,r)*P((n-r),r)
case:2 n-r<r then the total number of cases would be P(n,r)*[(C(n-r,1)*r + C(n-r,2)*(r-1)*2fact + C(n-r,3)*(r-2)*3fact-------
C(n-r,n-r)*(2r-n+1)*(n-r)fact
we got this expression by first arranging n objects in r places so we got P(n,r) AND then from remaining n-r objects we select 1 object and arrange in r places we got C(n-r,1)*r OR select 2 objects then arrange them in r places OR select 3 objects then arrange them in r places this will go upto n-r objects.
and we know that AND means multiplication OR means ADDITION
by this we will get the arrangement when no group will be vacant.
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Jitender