13
ДвҥїÑuÏ now in medical c
·2009-03-08 04:29:11
@ mathe...
are there exactly 2 on A, 2 on B??
or...atleast... 2 on A, 2 on B[7]
1
gordo
·2009-03-13 07:00:17
guyz..if the buks are the same, (i assume the event space to be only 2 buks in A and B), it leaves us with jus one case isint it...ie 2 buks in A 2 in B and 5 in C... and the sample space is 3^9...so probability = 1/39...in the second case if they are different...(ie distinct)...we choose 5 buks to be placed in C in 9C5ways...from the rest of the 4 buks we have to group them into 2 groups...which can be done in 4C2*2 ways..so total event space = 9C5*4C2*2 and the sample space being 39
so probability is ( 9C5*4C2*2)/(39)
11
Subash
·2009-03-13 06:14:15
Bhargav (or) mathematcian(or) b55:exactly
refer post 13 or do you want it with more than two
in my method x,y,z are the numbers of books on the three shelves
so obviously x+y+z=9
we find non negative integral solutions of this one that is 11C2 (i hope)
out of these total cases we need those in which there are 2 books on
A and B
that gives us two solutions (2,2,5) this is one and the other one is 2 interchanged
this is all i cud explain
if this makes completely no sense then im talking nonsense
62
Lokesh Verma
·2009-03-13 06:06:28
subash..
you can have more than 2 books on those shelves..
I think that is the mistake you are making..
I am not able to understand your method at all :(
11
Subash
·2009-03-12 06:51:06
i solved like
x+y+z=9
no of non negative solns is 11C2
out of these only 2 are required (2,2,5) [interchange of 2s give another solution]
so ans should be 2/11C2
where am i going rong please help
13
ДвҥїÑuÏ now in medical c
·2009-03-08 04:34:13
then... do u think bhaaiya is right???[7]
was that the ans????
13
ДвҥїÑuÏ now in medical c
·2009-03-05 03:14:09
what do u mean by books are different??
first one.. is it....28/729[7]
62
Lokesh Verma
·2009-03-07 23:10:01
answer wont change if you have same or different kinds of books.
62
Lokesh Verma
·2009-03-07 23:09:40
a+b+c=9
a>=2
b>=2
c>=0
a=x+2
b=y+2
c=z
so we need to solve
x+y+z=5
where x,y,z >0
so the number of solutions will be 7C2
You can find the probability by dividing by 39
11
Subash
·2009-03-06 04:05:28
yeah same prob in both cases
1
Ritika
·2009-03-06 00:04:24
d answer shud b d sem na....coz it's combinations acc. 2 me...not permutations.
13
ДвҥїÑuÏ now in medical c
·2009-03-05 23:31:15
i din get the second part....[2]
watz the ans of the first one????
39
Dr.House
·2009-03-05 03:19:35
answer is same in both cases?