MathCounts ProblemDate: 07/18/97 at 17:50:45 From: Wen Li Subject: Competition problem Hello Dr. Math, Here is a question that was on the 1995-96 MathCounts state competition. I have the answer, but I do not know how to get it. Two boys and four girls are officers of the Math Club. When the photographer takes a picture for the school yearbook, she asks the club's six officers and the faculty sponsor to sit in a row with the faculty sponsor in the middle and the two boys not next to one another. How many different seating arrangements are possible? I am sorry but I really have no idea on how to approach this problem. Could you please show me step by step the solution? Sincerely, Wen Li Date: 07/18/97 at 19:38:02 From: Doctor Anthony Subject: Re: Competition problem You have 3 seats either side of the faculty sponsor, and we need consider only the six seats of the officers. The number of arrangements of these six people with no restrictions on where they sit is 6! = 720. If the two boys are a and b, then if we group them as (ab) and move them about as a single entity, we have only 5 objects to arrange, and this can be done in 5! ways = 120 ways. However the boys could be paired as (ba) and again moved around as a single entity, giving a further 120 arrangements. So the total number of arrangements when the boys ARE together is 240. If now we subtract 240 from 720 we shall get the number of arrangements when they are NOT together. Number of arrangements with boys not together = 720 - 240 = 480 -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 07/19/97 at 17:40:49 From: Wen Li Subject: Wrong answer Hello Dr. Math, When I went back to check the answer for this problem, the answer was 528 arrangements, not 480. I asked Dr. Math this question because I wanted to learn the step by step way to answer this question. Could Dr. Anthony give me a detailed solution that will result in the answer being 528 arrangements? Thank you for your time. Sincerely, Wen Li Date: 07/20/97 at 05:25:28 From: Doctor Anthony Subject: Re: Wrong answer Yes. There are 48 extra arrangements with the boys not sitting together. These occur with the boys seated either side of the sponsor. With a on his left and b on his right, the 4 girls can be arranged in 4! = 24 ways. And now swapping a and b over so that a is on the right and b on the left of the sponsor, we get another 24 arrangements of the girls. This gives 24 + 24 = 48 extra arrangements with the boys not sitting together. Add this number to the 480 already found and you get 528. If you prefer, you can also get the answer by considering the possible arrangements by listing. In the diagram below, the O represent seats of the boys and * represents the seats of the girls. | is the seat of the sponsor. By the arguments given above, for any pair of seats occupied by the boys, there will be 48 arrangements of the girls. O * O | * * * the right hand boy could also occupy any seat to the right of where he is, giving 4 positions. With each of these there are a total of 48 arrangements of the girls. Total 4 x 48 * O * | O * * the right hand boy could occupy any seat to the right of his present position, giving 3 positions. With each of these there are 48 arrangements of the girls, giving a total of 3 x 48. * * O | O * * The right hand boy could occupy any seat to the right of his present position, giving 3 positions, and a further 3 x 48 arrangements. * * * |O * O There is no other position for the right hand boy, so this gives 1 x 48 arrangements. The total of all arrangements with the boys not sitting together is: (4 + 3 + 3 + 1) x 48 = 11 x 48 = 528. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/