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there are 7 teams A,B,C,D,E,F, each with 5 members. In how many ways can the 35 people be made to sit in a row such that every F team member sits next to i) at least one team G member 2^5*5!*30! (ANS 1.086*10^36) CORRECT

ii) exactly one team G member (ANS 4.7017*10^35) For the second one, I know you have to pair the 5 members of F and G into 5 couples and your new n is 30, each couple can be arranged in 2! ways and from my solution is (26C5)(25!)(2^5)*(5!) which is WRONG

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  • $\begingroup$ It really helps readability to format questions using MathJax (see FAQ). Regards $\endgroup$ – Amzoti Jun 11 '13 at 3:12
  • $\begingroup$ You're asking several questions and should break them apart. Also note that the answers you state are merely approximations, and not exact values. $\endgroup$ – Calvin Lin Jun 11 '13 at 3:21
  • $\begingroup$ oh dear: @azmoti wish I had time to actually learn jmax $\endgroup$ – iOSAndroidWindowsMobileAppsDev Jun 11 '13 at 3:48
  • $\begingroup$ @Calvin Lin The rounding should not be a problem to someone who knows their stuff. I forgot to mention that seat everyone else then seat the pairs in between them to satisfy the exactly part. I added the 26C5 for that. Please see edit $\endgroup$ – iOSAndroidWindowsMobileAppsDev Jun 11 '13 at 3:58
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(25!)(2^5)(26C5)(14400)

The missing part was actually choosing a representative for each pair from the two teams. This can be done in 5*5*4*4*3*3*2*2*1*1 (14400) ways

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  • $\begingroup$ I'm not sure how you're getting those numbers. Can you explain how you're counting configurations like "FGFABCFGF ..... "? $\endgroup$ – Calvin Lin Jun 11 '13 at 13:57

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