Problem Solving

12 persons sitting in a row,how select 4 out of them so that those who are adjecent will not be selected in any of the selection.

Sorry for posting it without option guys.I came across this model - and I wanted to know how this is approached.

Thanks in advance

Regards
Sai

objective is to select 4 out of 6 alternate seats.This can be done in 6C4 ways or 15 ways
Again this 6 alternate seats can be selected in 2C1 or 2 ways
so IMO ans is 2*15=30 ways

liferocks wrote:objective is to select 4 out of 6 alternate seats.This can be done in 6C4 ways or 15 ways
Again this 6 alternate seats can be selected in 2C1 or 2 ways
so IMO ans is 2*15=30 ways

great

thanks mate!

Hi,

Let us label the guys 1, 2, 3 and so on up to 12.

Case 1: Let 1 be the first member of the group.

1,3,5,_ -> 6 ways
1,3,6,_ -> 5 ways
1,3,7,_ -> 4 ways
1,3,8,_ -> 3 ways
1,3,9,_ -> 2 ways
1,3,10,_ -> 1 way

1,4,6,_ -> 5 ways
1,4,7,_ -> 4 ways
1,4,8,_ -> 3 ways
1,4,9,_ -> 2 ways
1,4,10,_ -> 1 ways

In a similar manner, 1,5,_,_ -> 4+3+2+1 = 10 ways
1,6,_,_ -> 3+2+1 = 6 ways
1,7,_,_ -> 2+1 = 3 ways
1,8,_,_ -> 1 way only.

Thus, if '1' is included, there are 21 + 15 + 10 + 6 + 3 + 1 ways (56 ways) of having 4 people.

Case 2: Let 2 be the first member of the group.

2,4,6,_ -> 5 ways
2,4,7,_ -> 4 ways

Thus, 2,4,_,_ -> 15 ways
2,5,_,_ -> 10 ways
2,6,_,_ -> 6 ways
2,7,_,_ -> 3 ways
2,8,_,_ -> 1 way

Thus, if '2' is included, there are 15 + 10 + 6 + 3 + 1 ways (35 ways) of having 4 people.

Case 3: If 3 is the first member of the group, there are 10 + 6 + 3 + 1 ways (20 ways) of having 4 people.

Case 4: If 4 is the first member of the group, there are 6 + 3 + 1 ways (10 ways) of having 4 people.

Case 5: If 5 is the first member of the group, there are 3 + 1 ways (4 ways) of having 4 people.

Case 6: If 6 is the first member of the group, there are 1 ways (1 way) of having 4 people.

Overall total = 56 + 35 + 20 + 10 + 4 + 1 = 126 ways.

4gmat_mumbai - it sounds right ..

I m just curious - is this model a common model in permutation combination or is it something new?

Because I have solved lots of problems in this topic - I was puzzled when I tried to solve this sum , though it sounded a regular model sum.

Yeah Mumbai's logic is sound, but really... would they give a problem like this... really? And doesn't this belong in the general area, not in the problem solving area?


I also initially thought it would be 2C1 x 6C4, but that really only seems to apply to permutations where you have to alternate two types of things (like boys and girls).