audi19 Posted September 1, 2007 Share Posted September 1, 2007 Is there a better way to understand combinations with constraints? For example, I'm reading the Manhattan GMAT prep Word Translations and I am trying to understand the explanation to this question but can't quite seem to get it. G, M, P, J, B, and C can sit next to each other in 6 adjacent seats. If M and J will not sit next to each other, how many different arrangements can the six people sit? Manhattan's explanation: 1) Find the number of ways in which six people can sit in 6 chairs: 6!= 720 2) Find the number of combinations in which M & J are sitting next to each other. a. There are 2! =2 ways in which M&J can be arranged and 4!=24 ways in which the other 4 can be arranged. b. Therefore there are 2x24=48 permutations for each seat pair. c. Since there are 5 seat pairs, there are 5x48=240 permutations in which M&J are sitting next to each other. 3) The number of permutations that they are not seated next to each other is 720-240 = 480. Is there a shortcut to this? Quote Link to comment Share on other sites More sharing options...
givemeanid Posted September 2, 2007 Share Posted September 2, 2007 I do not think there really is a shorter way of doing the above. Quote Link to comment Share on other sites More sharing options...
bekorchi Posted September 2, 2007 Share Posted September 2, 2007 My answer: that's the best way Quote Link to comment Share on other sites More sharing options...
audi19 Posted September 2, 2007 Author Share Posted September 2, 2007 I was afraid of that response. Thanks for your response! Quote Link to comment Share on other sites More sharing options...
regg Posted May 15, 2013 Share Posted May 15, 2013 I was afraid of that response. Thanks for your response! The easier approach would be to consider M and J as one unit. Now, consider the problem where there are 5 places and 5 ppl to be places in those five places. The number of permutations is 5! = 120. M and j although always beside each other can be arranged among themselves in 2!= 2 ways. So total permutations with M and J beside each other = 120 * 2 = 240. The answer is: Reduce 240 from total number of permutations of arranging 6 people on 6 places i.e. 6!=720 i.e. answer is 720-240 =480 Quote Link to comment Share on other sites More sharing options...
tucsonbella Posted July 10, 2013 Share Posted July 10, 2013 The easier approach would be to consider M and J as one unit. Now, consider the problem where there are 5 places and 5 ppl to be places in those five places. The number of permutations is 5! = 120. M and j although always beside each other can be arranged among themselves in 2!= 2 ways. So total permutations with M and J beside each other = 120 * 2 = 240. The answer is: Reduce 240 from total number of permutations of arranging 6 people on 6 places i.e. 6!=720 i.e. answer is 720-240 =480 This does seem like a faster approach. Thank you! Quote Link to comment Share on other sites More sharing options...
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