Raghu

• Jul 6th, 2005

 

For ex consider for A,B the combos are A,B and B,A i.e 2 combos are possible. 
 
For ex consider for A,B,C the combos (C,A,B),(A,C,B),(B,A,C) and (C,B,A) i.e 4 combos are  
possible. 
 
llly,for A,B,C,D the combos wld get doubled i.e 8 
 
for A,B,C,D,E the combos wld get doubled i.e 16 
 
for A,B,C,D,E,F the combos wld get doubled i.e 32 
 

nagu

• Sep 28th, 2005

 

correct

Nishant Joshi

• Nov 25th, 2005

 

the above sol is right but one combination is wrong ...

c cant come in middle of a and b so it shud be

(cab)(abc)(cba)(bac)

praveena

• Jan 31st, 2006

 

if we take second combination a,b,c. we can get 6 combinations i.e., (a,b,c),(a,c,b),(b,a,c),(b,c,a),(c,b,a),(c,a,b).

check it

sarfaraj

• Feb 23rd, 2006

 

In case of a , b , cc cant come in betn a & b as mention -side of a or b hence only 4 possible cases

dinesh

• Oct 16th, 2006

 

   ans is 240   

for a, b , c there are only 4 combinations since c cant come in b/w like wise d has 12 combinations

abcd  acbd bacd cabd bcad cbad

dabc  dacb dbac dcab dbca dcba ,

hence we can derive a simple eqn i.e for a, b,c=2! *2 =4

for a,b,c,d = 3! *2 =12

fora,b,c,d, e = 4! * 2 = 48

for a,b,c,d,e,f = 5!*2 =240

indu

• Oct 29th, 2006

 

there cant be six combinations with a,b,c coz c cant sit in between a and b

sandeep

• Jan 2nd, 2007

 

@ dinesh hey just want to correct you on ur answer of 12.... one of the condition mentioned is b wants to sit on any side of A and c takes seat after b so you cant have combinations that include acb where c comes in between a n b ... so there are just 8 arrangements and not 12 ... over all there are 32 such combinations.

chozha

 Profile Answers by chozha 

• Feb 16th, 2008

 

 it is clear that "c" cannot come in between a and b.so 4 combinations is right. 

pavan04kumar

 Profile Answers by pavan04kumar 

• May 25th, 2008

 

hiiiiiii

first B should be in the side of A so let us fix them as AB i.e they can be arranged = 2!=2

C shold be in the side of A or B ie (AB)C or C(AB) = 2! and A and B can be arranged among them in 2! ways from above  i.e total it can be arranged = 2! * 2! =4

now for D ABC are fixed as (ABC)D or D(ABC) so it can be arranged in = 4* 2! = 8
 
similarly for E= 8* 2! = 16

simlarly for F = 16* 2!= 32

so total no. of combinations after F is placed is 32 .... i suppose 32 as answer .... i am not sure abt my answer .... please correct me if i am wrong ....

a.sulaihabarvin

 Profile Answers by a.sulaihabarvin 

• Jan 4th, 2010

 

Answer is 32 only, because for 2 persons the possible combinations are 2,  and for 3 persons the possible combinations are 4, and so on for 6 members there are 32.

Persons->Combinations
2->2
3->4
4->8
5->16
6->32