Thanks a ton for this post…had a doubt about this for a long tym!…Just as an extension…for more complex polygons like regular hexagon, regular pentagon, etc. the same approach can be followed right?
For e.g. in a regular pentagon with 3 seats on each side, 15 ppl can be seated in 3*14! ways.
Chetan, in that case there will be 3 ways for the first person, true, but then there will be 11 chairs left and 9 people to place in them. So it will be 3 * 11P9.
sir, i am not able to understand how you are explaining the different view in case of isosceles triangle..?
Vinit, use your imagination here. Imagine you are the first person to arrive. Place yourself in each of the seats and ask yourself “how will the table look from here?”. You will find the answer to be slightly different in each of the 8 cases.
regards
J
sir, what is done for isosceles triangle, also holds true for rectangle, right? (referring to figure) because the person who 1st sits on a red seat, will be having two blue seats: one will be 1 position adjacent to him and other one at 4 positions far from him.. so all places are different here too..?
No Seema. Cut the figures out of cardboard and imagine yourself sitting at various positions round the table. Learn to visualise. You will see the same view from two positions in the rectangle.
In case there are 4 chairs on each side of an equilateral triangle, then no of ways= 4×15! or 3X15! as middle two positions will b same.
Hi J,
I am a regular reader of your posts and you are doing a wonderful job.
I have a doubt. After seeing the first example, the square one, my first instinct was 8p2*6p2*4p2. Why will this be wrong?
Because 8P2 would assume that you have selected a particular side to put 2 people at….And at the start of the problem, each of the 4 sides of the table is effectively identical. If you are the first to arrive, it doesn’t matter to you which side you sit at…
Sir I have a doubt. What if we have a square and and there are 8 chairs, 4 on vertexes and 4 on mid points of the side and have 8 people to sit ?
Then again answer will be 3*7 ? But here those 3 positions will include the vertex and its both its adjacent points ?
Hi J,
As in the above problem if there are total 15 people to be seated around an equilateral triangle with 3 people on the vertices and the rest as 4 on each side, then the total possible arrangement should be 5*14!? The 1st person’s options being the 4 types of seats on the sides + the type on the vertices. Hence total 5 options and the rest in 14! ways. Kindly confirm if the approach is right. Thank you. 🙂
Hi Sir, I recently came across a question in a mock, which was according to me similar to the question posted by apurba above. The explanation given for that question was not on the similar lines as you have explained above and for the similar reason the solution coming out it is different from what one expects after following the above approach. So i am in a dillema as to what the right approach is ? Question is : There is a table in the shape of an equilateral triangle as shown. In how many ways can 15 people be seated around this table ? Sorry for i cant show the diagram.. It is like 4 people on all three sides, all equidistant from each other and one peron on each of the three vertices. So in total 15 people.
Explanation provided for the question i posted above is as follows :
15 people can sit on 15 seats in 15! ways i.e. there are 15! different arrangements. But in any of these arrangements, if we shift all the people by 5 places or 10 places ahead clockwise, we get the arrangement which we have already considered in the 15! arrangements. Therefore, 15 people can sit around the table in 1/3 * 15! ways.
Sir J, Please enlighten why the solution is like above and why it is not 5 * 14! which is in accordance with your approach ? Please correct me if i am wrong somewhere !
sir, In rectangular table, it is similar to the question on circular arrangement on beads where if i cut through diagonal both of the arrangements look the same, then it should be divided by 2 right.? = 5*9!/2
No, not the same as beads, because we cannot turn it upside down. (Note that the answer is already taking care of that symmetry you mentioned – it can be perceived as 10!/2 = 5*9!)
In the isoceles triangle, won;t the same sides seats offer similar view( just as in rectangle, similar opposite sides had same view and same number of chairs, so we counted seats only on either side, so effectively 3 seats on same sides and 2 other, 5 and then 7!
Nope, try cutting an isosceles trying out of cardboard and checking for yourself by putting yourself in each position. The tip of the table will be in a different position / orientation for each seat.
CAT-holics 
Thanks a ton for this post…had a doubt about this for a long tym!…Just as an extension…for more complex polygons like regular hexagon, regular pentagon, etc. the same approach can be followed right?
For e.g. in a regular pentagon with 3 seats on each side, 15 ppl can be seated in 3*14! ways.
Absolutely. However if it is a non-regular figure then we would have to analyse it from scratch 🙂
regards
J
what about a case where there are 3 chairs on all sides of a rectangle. then the number of arrangements of 12 people will be 6*11! ?\
Effectively, yes. As long as the adjacent sides are noticeably different this will follow.
regards
J
what about 3 chairs on each side of square…with a total of 10 people to be seated..???will it be 3*9!???
Chetan, in that case there will be 3 ways for the first person, true, but then there will be 11 chairs left and 9 people to place in them. So it will be 3 * 11P9.
regards
J
wat about if 3 chairs on 2 side of equilateral triangle and 2 chairs on third side with a total of 8 people to be seated around???
Pranita, effectively again all the 8 positions will be different and so 8! would be my opinion.
regards
J
sir, i am not able to understand how you are explaining the different view in case of isosceles triangle..?
Vinit, use your imagination here. Imagine you are the first person to arrive. Place yourself in each of the seats and ask yourself “how will the table look from here?”. You will find the answer to be slightly different in each of the 8 cases.
regards
J
sir, what is done for isosceles triangle, also holds true for rectangle, right? (referring to figure) because the person who 1st sits on a red seat, will be having two blue seats: one will be 1 position adjacent to him and other one at 4 positions far from him.. so all places are different here too..?
No Seema. Cut the figures out of cardboard and imagine yourself sitting at various positions round the table. Learn to visualise. You will see the same view from two positions in the rectangle.
In case there are 4 chairs on each side of an equilateral triangle, then no of ways= 4×15! or 3X15! as middle two positions will b same.
No Sakshi, they won’t be the same; one will be towards the left and one towards the right of it’s respective side…
regards
J
Hi J,
I am a regular reader of your posts and you are doing a wonderful job.
I have a doubt. After seeing the first example, the square one, my first instinct was 8p2*6p2*4p2. Why will this be wrong?
Because 8P2 would assume that you have selected a particular side to put 2 people at….And at the start of the problem, each of the 4 sides of the table is effectively identical. If you are the first to arrive, it doesn’t matter to you which side you sit at…
regards
J
Why does the isosceles triangle example have an outcome equal to 2*7! ? As it is symmetrical
*why doesn’t….. *
Gaurav, can you find two seats which have EXACTLY the same view? 🙂
regards
J
Sir I have a doubt. What if we have a square and and there are 8 chairs, 4 on vertexes and 4 on mid points of the side and have 8 people to sit ?
Then again answer will be 3*7 ? But here those 3 positions will include the vertex and its both its adjacent points ?
Sorry for the typo sir. It should be 3*7! ?
No, it will be 2 * 7!, For the first person, all the vertices are identical.
regards
J
Hi J,
As in the above problem if there are total 15 people to be seated around an equilateral triangle with 3 people on the vertices and the rest as 4 on each side, then the total possible arrangement should be 5*14!? The 1st person’s options being the 4 types of seats on the sides + the type on the vertices. Hence total 5 options and the rest in 14! ways. Kindly confirm if the approach is right. Thank you. 🙂
Seems legit 🙂
regards
J
Thanks a lot J.
Hi Sir, I recently came across a question in a mock, which was according to me similar to the question posted by apurba above. The explanation given for that question was not on the similar lines as you have explained above and for the similar reason the solution coming out it is different from what one expects after following the above approach. So i am in a dillema as to what the right approach is ? Question is : There is a table in the shape of an equilateral triangle as shown. In how many ways can 15 people be seated around this table ? Sorry for i cant show the diagram.. It is like 4 people on all three sides, all equidistant from each other and one peron on each of the three vertices. So in total 15 people.
Explanation provided for the question i posted above is as follows :
15 people can sit on 15 seats in 15! ways i.e. there are 15! different arrangements. But in any of these arrangements, if we shift all the people by 5 places or 10 places ahead clockwise, we get the arrangement which we have already considered in the 15! arrangements. Therefore, 15 people can sit around the table in 1/3 * 15! ways.
Sir J, Please enlighten why the solution is like above and why it is not 5 * 14! which is in accordance with your approach ? Please correct me if i am wrong somewhere !
Thanks in advance. 🙂
1/3 * 15! is the same as 5 * 14! if you check… 🙂 Take care! Small mistakes kill you in CAT.
regards
J
Ohhhh i dint realize.. my bad 😦
Thanks Sir
sir,when the total of 9 people are seated in last qn with 3,3,3 then a total ways would be 6*8! because the identical 6 seats?
No. If it is an isosceles triangle, then 9! All seats will have a different view.
regards
J
sir, In rectangular table, it is similar to the question on circular arrangement on beads where if i cut through diagonal both of the arrangements look the same, then it should be divided by 2 right.? = 5*9!/2
No, not the same as beads, because we cannot turn it upside down. (Note that the answer is already taking care of that symmetry you mentioned – it can be perceived as 10!/2 = 5*9!)
regards
J
In the isoceles triangle, won;t the same sides seats offer similar view( just as in rectangle, similar opposite sides had same view and same number of chairs, so we counted seats only on either side, so effectively 3 seats on same sides and 2 other, 5 and then 7!
Nope, try cutting an isosceles trying out of cardboard and checking for yourself by putting yourself in each position. The tip of the table will be in a different position / orientation for each seat.
regards
J