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5 brothers 8 chairs

by vishal.pathak » Sun Nov 20, 2011 1:03 pm
In how many ways can 5 brothers sit in 8 chairs so that an elder brother is always to the right of a younger brother. (Assume there are no twins)
A) 4
B) 8
C) 56
D) 60
E) 120
OA not available

Related question: No of ways in which oldest is always next to the youngest
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by pemdas » Sun Nov 20, 2011 1:23 pm
i guess just simple combination not to leave the chairs empty 5C8=8!/(3!5!)=56
c
per related question we clue the oldest with the youngest and reduce the number of chairs, as far as no other constraint is given and consider permutation, 4P7=7!/4!=210
vishal.pathak wrote:In how many ways can 5 brothers sit in 8 chairs so that an elder brother is always to the right of a younger brother. (Assume there are no twins)
A) 4
B) 8
C) 56
D) 60
E) 120
OA not available

Related question: No of ways in which oldest is always next to the youngest
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by vishal.pathak » Sun Nov 20, 2011 1:35 pm
pemdas wrote:i guess just simple combination not leave the chairs empty 5C8=8!/(3!5!)=56
c
per related question we clue the oldest with the youngest and reduce the number of chairs, as far as no other constraint is given and consider permutation, 4P7=7!/4!=210
But wont the simple combination consider jumbling of the brothers. In combination 12345 is the same as 23145.
Here the requirement is that the oldest sits next his immediate younger brother (2nd agewise) followed by 3rd , 4th and then the youngest
So we have an order to be preserved here

Please clarify
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by pemdas » Sun Nov 20, 2011 1:57 pm
at first i meant all brothers sit side-by-side this would make only 4, but considering there are empty chairs to be filled we have 56

per your question, the order here and the order in perms/combos is different. In perms order means ABCDE is different from BACDE. In combos these two sets are identical, exactly what we need. Once the group of our brothers is arrayed (in age range) we have one group ABCDE or BACDE or CDEAB, doesn't matter we only need five of them to be in one group arrayed from the right to the left. We also must fill all empty chairs using different ways. Of course we cannot fill all the chairs with only five people but we will fragment through all 8 chairs in different ways within 56 ones.
vishal.pathak wrote:
pemdas wrote:i guess just simple combination not leave the chairs empty 5C8=8!/(3!5!)=56
c
per related question we clue the oldest with the youngest and reduce the number of chairs, as far as no other constraint is given and consider permutation, 4P7=7!/4!=210
But wont the simple combination consider jumbling of the brothers. In combination 12345 is the same as 23145.
Here the requirement is that the oldest sits next his immediate younger brother (2nd agewise) followed by 3rd , 4th and then the youngest
So we have an order to be preserved here

Please clarify
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by kanwar86 » Sun Nov 20, 2011 2:02 pm
vishal.pathak wrote:
pemdas wrote:i guess just simple combination not leave the chairs empty 5C8=8!/(3!5!)=56
c
per related question we clue the oldest with the youngest and reduce the number of chairs, as far as no other constraint is given and consider permutation, 4P7=7!/4!=210
But wont the simple combination consider jumbling of the brothers. In combination 12345 is the same as 23145.
Here the requirement is that the oldest sits next his immediate younger brother (2nd agewise) followed by 3rd , 4th and then the youngest
So we have an order to be preserved here

Please clarify
IMO, Pemdas has only selected the arrangements 5C8=8!/(3!5!)=56 (and not permuted them).
Suppose we have 5 chairs and 5 brothers. We can arrange these 5 persons in 5! ways. However, there is only one way we can select an arrangement in which each elder brother is sitting next to a younger brother.
It's just now we have 8 chairs instead of 5. This implies we have more selections. Still, for each selection (though number of arrangements exist), there is only one arrangement which follow the given constraint. Hence, only by finding number of selections (56 in this case), we can find the solution.
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by vishal.pathak » Sun Nov 20, 2011 2:18 pm
pemdas wrote:at first i meant all brothers sit side-by-side this would make only 4, but considering there are empty chairs to be filled we have 56

per your question, the order here and the order in perms/combos is different. In perms order means ABCDE is different from BACDE. In combos these two sets are identical, exactly what we need. Once the group of our brothers is arrayed (in age range) we have one group ABCDE or BACDE or CDEAB, doesn't matter we only need five of them to be in one group arrayed from the right to the left. We also must fill all empty chairs using different ways. Of course we cannot fill all the chairs with only five people but we will fragment through all 8 chairs in different ways within 56 ones.
So then is the question about arrangement of chairs for the brothers. Are the 8 chairs themselves not fixed. Are we saying that it is possible to arrange the chairs in different order

I thought that the chairs are themselves fixed too, I thought that the chairs are queued too and in these queued chairs we have to make the brothers sit in a specific order and the no of arrangements will depend on the number of empty chairs between brothers

Like for example if o stands for occupied chair and | stands for empty chair then I thought that the arrangements will be like
ooo||oo|
oooo|o||
etc

What will be the number of arrangements in such a case where bothers have to sit in the order of oldest to youngest and chairs are numbered from 1 to 8, kept in a queue and cannot be moved
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by pemdas » Sun Nov 20, 2011 2:47 pm
chairs are not moved, people are moved through chairs; the chairs are fixed
with combos and perms when novice, it's very tempting to misapply nCm for mCn
one rule here when you have to arrange the less number amongst big number, like 5 brothers among 8 chairs you will fragment some places (leave them empty and allow for some space between your arranged units, here brothers). If you had 8 brothers and 5 chairs you would have to miss some brothers in each successive arrangement way.
>>don't throw your quickest at combos and perms, they require day-long thinking till one's accustomed; later it's coming in a moment
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by shankar.ashwin » Sun Nov 20, 2011 9:07 pm
Another way of looking at it. We have 8 chairs here,

_ _ _ _ _ _ _ _ and 5 people to occupy these 8 chairs.

Now we have 8 chair and 5 to pick (to seat the 5 brothers) is 8C5 = 56.

These include selecting different chair numbers such as (1,2,3,4,5) (1,2,4,5,8) etc.

Now after we have selected our 5 chairs, there is only 1 way of seating the 5 people (i.e in ascending order of their age). So in total we have 8C5*1 = 56.

PS. If the age restriction was not mentioned, the 5 brother can sit among themselves in 5! ways, so then total would be 8C5*5!.

Hope this makes sense.

@Pemdas Is 5C8 a valid combination expression? In nCr, we always have n>r. 5C8 would be 5!/(-3!)*8!
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by HSPA » Sun Nov 20, 2011 9:53 pm
Let the chairs be numbered 1 to 8

Eldest: He can choose among 'only 4' chairs (8,7,6,4) = 4c1 = 4
second eldest: He can choose only 4 chairs (7,6,5,4) = 4
middle fellow: 6,5,4,3
second youngest: leave 3 on left and 1 on right: 4
similary youngest has only last 4 chairs to select.(4,3,2,1)

I got 4x4x4x4x4 < number of ways of allocating 8 chairs to 5 people 8p5..
what is the blunder in my approach
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by kanwar86 » Sun Nov 20, 2011 11:48 pm
HSPA wrote:Let the chairs be numbered 1 to 8

Eldest: He can choose among 'only 4' chairs (8,7,6,4) = 4c1 = 4
second eldest: He can choose only 4 chairs (7,6,5,4) = 4
middle fellow: 6,5,4,3
second youngest: leave 3 on left and 1 on right: 4
similary youngest has only last 4 chairs to select.(4,3,2,1)

I got 4x4x4x4x4 < number of ways of allocating 8 chairs to 5 people 8p5..
what is the blunder in my approach
IMO, We usually go for the solution you just described when we need to permute to find the number of ways for a particular thing to occur. Now, we have a constraint (that the elder brother is sitting to the right of younger brother). If you watch carefully, you can see a number of cases in which this condition has been violated.
Let's check your solution
Eldest E1: He can choose among 'only 4' chairs (8,7,6,4) = 4c1 = 4.....(8,7,6,4)
second eldest E2: He can choose only 4 chairs (7,6,5,4) = 4............(7,6,5,4)
middle fellow E3: 6,5,4,3..............................................(6,5,4,3)
second youngest E4: leave 3 on left and 1 on right: 4...................(5,4,3,2)
similary youngest E5 has only last 4 chairs to select.(4,3,2,1).........(4,3,2,1)
Now if (E1, E2, E3, E4, E5) are sitting in (4,5,6,2,3) clearly it does not satisfy the given condition.
Similarly you can find other arrangements. That's why this method cannot be used here.
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by pemdas » Mon Nov 21, 2011 12:00 am
shankar.ashwin wrote: @Pemdas Is 5C8 a valid combination expression? In nCr, we always have n>r. 5C8 would be 5!/(-3!)*8!
i always put this in different order since attending to this forum, i meant 8C5 as you see from formula that followed. Thanks for correcting me, i should remember the formula look not to misguide others.
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