|
Author
|
Message
|
![[Post New]](/templates/default/images/icon_minipost_new.gif)
26 Dec 2006 12:37:38 IST
|
|
|
2 * 8 C = 16 7
|
this reply:
0 points
(with 0
in
0
votes
)
[?]
|
|
You have to be logged on to rate
|
|
|
29 Dec 2006 12:56:23 IST
|
|
|
PERMUTATION IS NOT REQUIRED HERE,ONLY COMBINATION CONCEPT IS TO BE USED SINCE WE HAVE TO SELECT 7 PEOPLE AND NOT TO ARRANGE THEM. TO SELECT 7 PEOPLE OUT OF 16 SO THAT NO TWO ARE COMSECUTIVE IS = 9 C 7
|
this reply:
5 points
(with 1
in
1
votes
)
[?]
|
|
You have to be logged on to rate
|
|
|
29 Dec 2006 15:40:38 IST
|
|
|
to select 7 people such that none is consecutive would be
First person can occupy any of the 16 chairs
Second person can sit in rest of the 13 chairs
Third in 11, Fourth in 9,Fifth in 7, Sixth in 5 ,proceeding in this way
Therefore total number of ways to sit= 16*13*11*9*7*5*3=2162160
|
The yardstick of human intelligence is the ability to overcome the last fallacy |
this reply:
2 points
(with 0
in
1
votes
)
[?]
|
|
You have to be logged on to rate
|
|
|
29 Dec 2006 16:45:33 IST
|
|
|
there are 16 people and 7 have 2 b seleced there r 9 people left and as the 7 people have not to sit consecutivel so choices 10p7 (as 10 gaps) 10!/3! 604800.
|
this reply:
0 points
(with 0
in
0
votes
)
[?]
|
|
You have to be logged on to rate
|
|
|
29 Dec 2006 16:58:17 IST
|
|
|
seven people can sit in 8 places as no two persons should be consecutive.
no. of ways=8C7*7P7*2
|
this reply:
0 points
(with 0
in
0
votes
)
[?]
|
|
You have to be logged on to rate
|
|
|
29 Dec 2006 17:00:15 IST
|
|
|
10C7 120
|
this reply:
0 points
(with 0
in
0
votes
)
[?]
|
|
You have to be logged on to rate
|
|
|
29 Dec 2006 18:03:11 IST
|
|
|
Suppose that 7 people are sitting round the table(and these are the chosen people)
Make other seven out of nine left to sit between each two.
One will sit anywhere he want.
For, each place where the last one will sit give a different arrangement.
Hence the answer is four due to circular symmetry.
this is the solution, I hope, if all the people are identical
if all the people are different, then this solution can be extended to C(16,7)*C(9,7)*{C(7,2)+C(7,1)}
Anyone has a doubt??
|
Imagination is more important than knowledge
-------Albert Einsetein |
this reply:
0 points
(with 0
in
0
votes
)
[?]
|
|
You have to be logged on to rate
|
|
|
31 Dec 2006 10:15:38 IST
|
|
|
Since we have to select 7 people out of 16 so that no two are consecutive:- Suppose we take one and select it as 16C1! Now the two immediate to it can't be selected ! Left are 7 which can be selected as they are alternate! therefore we have to select 6 out of them as we have already selected ONE! therefore 7C6 Answer is 16C1 * 7C6 = 112
|
this reply:
0 points
(with 0
in
0
votes
)
[?]
|
|
You have to be logged on to rate
|
|
|
31 Dec 2006 13:39:11 IST
|
|
|
9C7 must b the answer .we have 16 seats around a table(round). now it can b thought of as we have to select 7 seats out of the 9 gaps created by seating the remaining 9. here permutation is not required as persons r fixed on their seats.
|
keep it up!!!!!!!!!11 |
this reply:
0 points
(with 0
in
0
votes
)
[?]
|
|
You have to be logged on to rate
|
|
|
14 Jan 2007 20:18:06 IST
|
|
|
727
|
this reply:
0 points
(with 0
in
0
votes
)
[?]
|
|
You have to be logged on to rate
|
|
|
16 Jan 2007 15:33:11 IST
|
|
|
36
|
this reply:
0 points
(with 0
in
0
votes
)
[?]
|
|
You have to be logged on to rate
|
|
|
17 Jan 2007 19:32:43 IST
|
|
|
1,3,5,7,9,11,13,
|
this reply:
0 points
(with 0
in
0
votes
)
[?]
|
|
You have to be logged on to rate
|
|
|
18 Jan 2007 12:52:02 IST
|
|
|
it iis arr 7 vacancies &9 people on round table such that no 2 vacancies r 2gether 8! * (8c7) * 6!
|
this reply:
0 points
(with 0
in
0
votes
)
[?]
|
|
You have to be logged on to rate
|
|
|
20 Jan 2007 19:51:16 IST
|
|
|
swordfish #2 result The solution given by ravitej is correct he deserves a prize
|
this reply:
0 points
(with 0
in
0
votes
)
[?]
|
|
You have to be logged on to rate
|
|
|
21 Jan 2007 22:22:55 IST
|
|
|
total no of people to be arranged arround a circular table = 16 no. of ways in which remaining 9 people may be arranged = ( 9 - 1)! = 8! the other 7 persons may be arranged among themselves in 7! ways. Thus total no. of ways = 8! * 7! = 203212800
|
this reply:
0 points
(with 0
in
0
votes
)
[?]
|
|
You have to be logged on to rate
|
|
|
21 Jan 2007 23:06:52 IST
|
|
|
16 7*7!/2!
|
this reply:
0 points
(with 0
in
0
votes
)
[?]
|
|
You have to be logged on to rate
|
|
|
23 Apr 2007 16:34:03 IST
|
|
|
hey i need ans. fr those ques. askd and also i want to be informed fr nxt contest....i wil prepare fr it
|
this reply:
0 points
(with 0
in
0
votes
)
[?]
|
|
You have to be logged on to rate
|
|
|
14 May 2007 10:51:43 IST
|
|
|
16*7
|
     
----------------------------------------------------
simpler@INDIAN |
this reply:
0 points
(with 0
in
0
votes
)
[?]
|
|
You have to be logged on to rate
|
|
|
14 May 2007 10:58:54 IST
|
|
|
hi friends whr c'ld I find bitsat sample papers
|
----------------------------------------------------
simpler@INDIAN |
this reply:
0 points
(with 0
in
0
votes
)
[?]
|
|
You have to be logged on to rate
|
|
|
11 Jan 2008 22:22:19 IST
|
|
|
since we have to select 7 ppls frm 16 but no consicutives shoul b selected then
lets start this way
consider a person as the head...
now including the head we have 8 members in which no two are consicutive..
there4 the total no. of ways 4 selection of 7 ppl frm 8 is
8c7
now the person just next to the head is excluded in the 1st case ,so now lets count him now and the other 7 ppl who were left before , so now again we have 8 members and no 2 of them are consicutive
again the total no of ways to select 7 ppl frm 8 is
8c7
now the case 1 is independable of the case 2 there4
total no of cases will be
8c7+8c7=2*8c7=2*8c1=16
ans is 16
|
this reply:
0 points
(with 0
in
0
votes
)
[?]
|
|
You have to be logged on to rate
|
|
|
|
|
Moderation Team
Community Discussion Question:
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
3
