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January 29th, 2016, 05:06 AM  #1 
Senior Member Joined: Oct 2013 From: Far far away Posts: 422 Thanks: 18  permutation combination problem
In how many ways can the numbers 1, 2, 3, 4, 5, 6, 7 be arranged so that A) 1 and 7 are adjacent? B) 1 and 7 are not adjacent? C) 1 and 7 are exactly 3 spaces apart? My attempt: Order of arrangement matters with numbers > 21 is different from 12 A) we can consider the 7digit number as slots to fill e.g. _ _ _ _ _ _ _ If we group the slots as a pair of adjacent slots there are 6 such pairs. Since order matters (17 is distinct from 71) there are 2*6 = 12 different arrangements of 1 and 7 adjacent to each other (not considering the other five slots). Now there are 5 slots open. The number of different arrangements possible with these is 5! Therefore, the number of arrangements where 1 and 7 are adjacent = 12*5! B) there are seven slots to fill as in _ _ _ _ _ _ _ Scenario a: 1_ _ _ _ _ _ in this case there are 5 possible slots where 1 and 7 are not adjacent. Then the remaining 5 slots can be filled in 5! ways giving us 5*5! different permutations. Now replace 7 for 1 in the above scenario and we have a total of 5*5!*2 total permutations. Scenario b: _1_ _ _ _ _ in this case there are 4 slots for 7 to go in. For each of these arrangements there's 5! arrangements in the remaining slots. Swap 1 and 7 and that gives us a total number of arrangements = 4*5!*2 And so on... We get total number of possible arrangements where 1 and 7 aren't adjacent = 2(5*5!*2) + 5(4*5!*2) C) possibilities 1_ _ _ 7_ _ ; _1_ _ _ 7_ ; _ _1_ _ _ 7 and swap 7 and 1 we get 3*2 arrangements with 1 and 7 separated by 3 spaces. Now the remaining 5 slots can be filled in 5! ways giving us a total of 3*2*5! different arrangements where 1 and 7 are separated by 3 spaces. Am I correct? Is there a better/different way to solve this problem? Thanks 
January 29th, 2016, 09:32 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 896 
Here is how I would do A since you want 1 and 7 to be together treat "17" as a single "letter". Then there are 6 letters and there are 6! ways to order them. Treating "71" as a single letter gives the same thing. So there are 2(6!)= 2(720)= 1440, not 4*5!*2= 8*120= 960.
Last edited by Country Boy; January 29th, 2016 at 09:35 AM. 
January 30th, 2016, 01:29 AM  #3  
Senior Member Joined: Oct 2013 From: Far far away Posts: 422 Thanks: 18  Quote:
And also you haven't considered that there are 4 possible locations for ''17'' or ''71'' as in 17_ _ _ _ _ _ 17_ _ _ _ _ _ 17_ _ _ And so on... Am I right? Last edited by shunya; January 30th, 2016 at 01:34 AM.  
January 30th, 2016, 02:03 AM  #4 
Math Team Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 
No. What CountryBoy means is that you group the 1 and the 7 together and treat them as one block. So your new list of objects would be 2, 3, 4, 5, 6, 17 (or 71, depending on which order the numbers are in.) Now you have 6 objects to arrange. There are 6! = 720 ways to arrange them. There are also two ways to order the 1 and 7. Thus there are a total of 2(6!) = 1440 arrangements. It gives the same answer as your method. For part B), there is an easier way. There are 7! = 5040 ways to arrange the numbers with no restrictions and from part A), there are 1440 ways to arrange the numbers with 1 and 7 adjacent. Thus, the number of ways to arrange the numbers so that 1 and 7 are not adjacent is 5040  1440 = 3600 ways. Your answer is much too large. It's more than the number of arrangements with no restrictions. C) looks fine. 

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