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Difficult Math Problem #99 - Permutations

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Difficult Math Problem #99 - Permutations

by 800guy » Mon Feb 19, 2007 8:15 am
Among 5 children there are 2 siblings. In how many ways can the children be seated in a row so that the siblings do not sit together?
(A) 38
(B) 46
(C) 72
(D) 86
(E) 102

oa coming when some people respond with explanations. from diff math doc
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by Neo2000 » Mon Feb 19, 2007 8:51 am
The 5 kids can be arranged among themselves in 5! ways

Consider the 2 siblings as 1. Then 4 kids can be arranged among themselves in 4! x 2! ways ( since the 2 can be arranged among themselves in 2! ways)

So Total Number of ways in which 5 kids can be arranged so that 2 particular kids DO NOT sit together = 5! - (4! x 2!)

Answer Option (C)
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Re: Difficult Math Problem #99 - Permutations

by banona » Mon Feb 19, 2007 1:06 pm
800guy wrote:Among 5 children there are 2 siblings. In how many ways can the children be seated in a row so that the siblings do not sit together?
(A) 38
(B) 46
(C) 72
(D) 86
(E) 102

oa coming when some people respond with explanations. from diff math doc
Do you mean two couples of siblings or two kids who are siblings ?
I appologize for my Frenchy-English.
I am working on it.
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by aim-wsc » Mon Feb 19, 2007 6:47 pm
@banona well this is all you have :)
work on it.

by the way you are doing good job at math section.
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Re: Difficult Math Problem #99 - Permutations

by 800guy » Mon Feb 19, 2007 6:51 pm
banona wrote:
800guy wrote:Among 5 children there are 2 siblings. In how many ways can the children be seated in a row so that the siblings do not sit together?
(A) 38
(B) 46
(C) 72
(D) 86
(E) 102

oa coming when some people respond with explanations. from diff math doc
Do you mean two couples of siblings or two kids who are siblings ?
this isn't clearly written perhaps, but i was thinking two kids who are siblings.
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Re: Difficult Math Problem #99 - Permutations

by Neo2000 » Mon Feb 19, 2007 6:53 pm
banona wrote: Do you mean two couples of siblings or two kids who are siblings ?
banona,
I believe you are misinterpreting the question. The Q only says 2 siblings NOT 2 sets of siblings. You require atleast 2 children to have a sibling pair.
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Re: Difficult Math Problem #99 - Permutations

by 800guy » Mon Feb 19, 2007 6:54 pm
Neo2000 wrote:
banona wrote: Do you mean two couples of siblings or two kids who are siblings ?
banona,
I believe you are misinterpreting the question. The Q only says 2 siblings NOT 2 sets of siblings. You require atleast 2 children to have a sibling pair.
good eye, neo. i guess the prompt is pretty clear...
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Re: Difficult Math Problem #99 - Permutations

by banona » Tue Feb 20, 2007 2:38 pm
Neo2000 wrote:
banona wrote: Do you mean two couples of siblings or two kids who are siblings ?
banona,
I believe you are misinterpreting the question. The Q only says 2 siblings NOT 2 sets of siblings. You require atleast 2 children to have a sibling pair.
.

thank you for the your remark, but you would undestand that I am struggling againt my frenchy-english and subtile stuffs like those usually let me confused,

thanks again
I appologize for my Frenchy-English.
I am working on it.
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by banona » Tue Feb 20, 2007 3:12 pm
Answer must be C,
wHY ?
let consider the following row ;

A B C D E
If five students are to be arranged so that one siblings (2kids) do never sit togoether, so
the only possibilities are for the siblings to sit on the followings positions ;

A C
A D
A E
B D
B E
C E

we have at all 6 configurations,
for each configuation, the two sibilings can switch mutually theirs positions, so that it leads to 12 configurations in sum,

for each one of these 12 configurations, the other children can sit a number of ways equal to 3! ( WHICH IS THE PEMUTATIONS OF THE
OTHER CHILDREN AROUND THE VACCANT POSITIONS)

Then, the total ways are : 12 * 3! = 72

I believe answer is C.
I appologize for my Frenchy-English.
I am working on it.
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by banona » Tue Feb 20, 2007 3:15 pm
aim-wsc wrote:@banona well this is all you have :)
work on it.

by the way you are doing good job at math section.

Thank you aim-wsc,
I hope I will so on the 5th March,
My God, it is toooooo close !!!!!!!!!
I appologize for my Frenchy-English.
I am working on it.
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oa

by 800guy » Wed Feb 21, 2007 9:15 am
oa:

The total number of ways 5 of them can sit is 120
when the siblings sit together they can be counted as one entity
therefore the number of ways that they sit together is 4!=24, but since
the two siblings can sit in two different ways e.g. AB and BA we multiply 24 by 2 to get the total number of ways in which the 5 children can sit together with the siblings sitting together - 48
In other words 4P4*2P2
the rest is obvious 120-48=72
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by BTGmoderatorRO » Sun Nov 05, 2017 10:55 am
we want to arrange 5 children in a row, in a way that 2 siblings do not sit together.
total number of ways of arranging the children is = 5! =120ways
if we are to arrange the children in a way that the two siblings MUST sit together, we can do that in the following way:
first we count the two siblings as one entity, so that we are left with four entities to arrange. This can be done in 4! way s. The siblings themselves can be arranged in 2! ways.
so number of possible arrangements, n, is
n= 4! * 2!
n= 48 ways
therefore, the number of arrangements possible, if the two siblings must not sit together.
= total possible arrangement - n
= 120- 48 ways
=72ways
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by Scott@TargetTestPrep » Tue Oct 29, 2019 6:03 am
800guy wrote:Among 5 children there are 2 siblings. In how many ways can the children be seated in a row so that the siblings do not sit together?
(A) 38
(B) 46
(C) 72
(D) 86
(E) 102
If there are no restrictions, 5 people can be seated 5! = 120 in a row. Now let's assume that the 2 siblings must sit together, then the number of ways this can be done is 4! x 2 = 24 x 2 = 48. However, since they actually can't sit together, the number of ways they can't sit together is 120 - 48 = 72.

Answer: C

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