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18 Nov 2010, 22:45
Kudos
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This question is not from any book or any test, this is a real time combination problem -
There is a group of 5 [ABCDE] people 3 boys [ABC] and 2 girl[DE]. I need to find out the combination to arrange these 5 people on 5 seats, in a manner that
1- out of 3 boys, 2 have to sit on the corner. In other words, both corner can be occupied by the boys.
2- One of the boy [Let say C] can't not sit with the girl [D]. (Because they have some personal issues).
I thought of posting this question because not only we all are going for a Movie and we have to create possible combination but also I am very weak in combinations.
There is a group of 5 [ABCDE] people 3 boys [ABC] and 2 girl[DE]. I need to find out the combination to arrange these 5 people on 5 seats, in a manner that
1- out of 3 boys, 2 have to sit on the corner. In other words, both corner can be occupied by the boys.
2- One of the boy [Let say C] can't not sit with the girl [D]. (Because they have some personal issues).
I thought of posting this question because not only we all are going for a Movie and we have to create possible combination but also I am very weak in combinations.
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19 Nov 2010, 07:58
Kudos
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zerotoinfinite2006
Easiest way to deal with combination is to first assume the situation is simple i.e. 5 people arranged in a row is in 5! ways.
Now bring in a complication - The corners should be occupied by boys.
So you could just choose two boys, make them sit at the corners in 2! ways and arrange the rest of the 3 people.
But, another complication - A boy and a girl cannot be together. Cannot be together is handled by finding out the ways in which they can be together and then subtracting from the total.
The problem is that if the boy who cannot be together with the girl is sitting at a corner, the situation is different. If he is sitting in one of the middle 3 seats, the situation is different.
Lets say 3 Boys are B1, B2, B3 and 2 girls are G1, G2. B1 and G1 do not want to sit together.
Two cases:
Case 1: B1 sits in one of the middle seats
So B2 and B3 sit on the corner seats in 2! ways.
G1, G2 and B1 need to sit in the middle seats and B1, G1 should not be together. Then G2 must sit between them. So 2 ways: B1, G2, G1 and G1, G2, B1
Total number of ways here = 2! * 2 = 4 ways
Case 2: B1 sits in one of the corner seats
Select one out of B2 and B3 to sit on the other corner in 2C1 = 2 ways
B1 and the other guy can sit on two corners in 2! ways.
G1 has only 2 places available for her (e.g. B1 * * * B2 so she cannot take the second place in the row. She must take the 3rd or 4th place) so choose 1 out of these 2 places in 2C1 = 2 ways
The other two people can arrange themselves in the other 2 vacant spots in 2! ways
Total number of ways in this case = 2 * 2! * 2 * 2! = 16 ways
Total number of ways combined = 4 + 16 = 20 ways
Note: Generally when working with 'two people should not sit together', we find out the number of ways in which they can sit together and then subtract it from the total. Here, since number of people is very small, its just easier to imagine it and directly solve.
19 Nov 2010, 09:06
Kudos
Bookmarks
I tried the "glue" method.
Total number of ways to arrange 5 people with 2 of the 3 boys in the corners = 3C2 * 2* 3! = 3 * 12 = 36 (you need to multiply by 2 as you need to count B1(left extreme)---B2(right extreme) again as B2---B1; its a permutation scenario).
Now glue together D and E, your friends who have issues, and make them sit together always (they may become friends
!).
Scene 1: They sit together in last 2 seats to the right with the boy occupying the corner seat.
Others can now sit in 2!*2!*1! = 4 ways (extreme left seat can only be taken by one of the 2 boys).
Similarly, D and E can sit together on the first 2 seats with the boy taking the extreme left/corner seat.
So Scene 1 can have 4*2 = 8 outcomes.
Scene 2: They sit together in the 2nd and 3rd or 3rd and 4th seats (seat order counting from left to right)
For 2nd and 3rd the arrangements are: 2*2*1 = 4 (the extreme left and extreme right can only be taken by the other 2 boys - so first seat on the left can be taken in 2 ways; then D and E can sit in reverse order of E and D so we need a second 2. The other girl can sit in the other non-corner seat). For the 3rd and 4th seats it will be the same outcome.
So Scene 2 can have 4*2 = 8 outcome.
Required arrangements = Total ways - 'Glued together' ways = 36-[8+8] = 36-16 = 20 arrangements.
Total number of ways to arrange 5 people with 2 of the 3 boys in the corners = 3C2 * 2* 3! = 3 * 12 = 36 (you need to multiply by 2 as you need to count B1(left extreme)---B2(right extreme) again as B2---B1; its a permutation scenario).
Now glue together D and E, your friends who have issues, and make them sit together always (they may become friends
!).Scene 1: They sit together in last 2 seats to the right with the boy occupying the corner seat.
Others can now sit in 2!*2!*1! = 4 ways (extreme left seat can only be taken by one of the 2 boys).
Similarly, D and E can sit together on the first 2 seats with the boy taking the extreme left/corner seat.
So Scene 1 can have 4*2 = 8 outcomes.
Scene 2: They sit together in the 2nd and 3rd or 3rd and 4th seats (seat order counting from left to right)
For 2nd and 3rd the arrangements are: 2*2*1 = 4 (the extreme left and extreme right can only be taken by the other 2 boys - so first seat on the left can be taken in 2 ways; then D and E can sit in reverse order of E and D so we need a second 2. The other girl can sit in the other non-corner seat). For the 3rd and 4th seats it will be the same outcome.
So Scene 2 can have 4*2 = 8 outcome.
Required arrangements = Total ways - 'Glued together' ways = 36-[8+8] = 36-16 = 20 arrangements.
