IIT JEE , AIEEE Forum - Mathematics , Algebra

sharmadon

20 students have been seated along a long table, 10 on each side. 8 students are selected at random.then what is the probability that 4 students belong to one side and 4 to the other side but no two of them are adjacent to each other?

computer001

edit:pls refer below

armaan

^{is it ...20}C₈/2*⁵C_4...?

armaan

sumbody confirm my answer ....

LAMPARD

I am getting it as (1/672)².
There can be 120 groups of 4 students selected from 10 students in such a way that are not adjacent to each other.Total number of groups of 4 is 5040.
Thus probability that the selected group of students are not adjacent to each other is 120/5040=1/42.
Now,the probability that the 4 students should belong to the 1st group is 1/16.
Thus,total probability for group of 1st 4 students is (1/16)*(1/42)=1/672.
The same applies for the 2nd group of 4 students.Thus total probability is (1/672)².
Am i correct???

computer001

EDIT: refer bottom post

LAMPARD

computer001..
20 is only for the 1st person in the row in your case.
See,let us name them as A,B,C,D,E,F,G,H,I,J.
Now,A has pairs[A,C,E,s] where s belongs to {G,H,I,J}.Thus,there are 4 pairs in this case.
Another one for A is [A,D,F,s] where s belongs to {H,I,J}.Thus,in this case,3 groups.
Continuing like this,you get 20 pairs beginning with A itself.Now,consider the same with B.You will get only 10 such pairs for B.Thus A and J have 20 pairs each beginning with them while rest 8 have 10 such.Thus,totally,there are 120 such pairs.Same goes for 2nd group of 10 students.

computer001

no lampard u r mistaken...
pls refer to the idea of recurssions...
im takin every case into account

LAMPARD

But how?It is clearly evident that there are 120 cases.Can u explain the idea of recursions??

akhil_o

yeah we find it for 4 ppl out of 10 and square it,
and then multiply by the Probability that they r sitting 10 and 10 across

sorry didnt see above post:

what r recursions?

computer001

im sooo sorry it shud be:
{4+3+2+1 + 3+2+1 +2+1 +1} +{3+2+1 +2+1 +1} +{2+1 +1} + {1}
= 20+10+4+1=35
so v have 35*35/20C8

IDREAMIIT

----------sry did nt read the qs properly--------------

armaan

ok tell me the mistake i ll tell u sets ...

let [1,2,3,4,5,6,7,8,9,10] be the students

select them

[1,3,5,7] [1,3,5,8,] [1,3,5,9] [1,3,5,10] [1,3,6,8] [1,3,6,9] [1,3,6,10] [1,4,6,8] [1,4,6,9] [1,4,6,10] [1,4,7,9] [1,4,7,10] [1,4,8,10] [,1,5,7,9] [1,5,7,10] [1,5,8,10] [1,6,8,10][2,4,6,8] [2,4,6,9] [2,4,6,10] [2,4,7,9] [2,4,7,10] [2,5,7,9] [2,5,7,10] [2,5,8,10] [2,6,8,10] [3,5,7,9] [3,5,7,10] [3,5,8,10] [3,6,8,10] [4,6,8,10]

31 caeses...

so answer is 31*31/¹⁰C^_8..

correct me if m rong... how did u form 35 cases ??

armaan

ahh shit sorry ...

its 31*31/²⁰C₈

IDREAMIIT

edited

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