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Consecutive Numbers and Subset Choice (Posted on 2016-02-09) Difficulty: 3 of 5
M is the number of six-element subsets that can be chosen from the set of the first 15 positive integers so that at least three of the six numbers are consecutive.
Find M.

No Solution Yet Submitted by K Sengupta    
Rating: 3.0000 (1 votes)

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Solution possible (?) solution | Comment 4 of 6 |

Lets call a subset of ( 1,2,3,4…15)  having 6 members a 6-subset.

  1. All of the 6-subsets with at least one triplet of consecutive integers must contain at least one   multiple of 3.

    b. The quantity of  6-subsets with no triplet of consecutive integers is therefore C(10,6).      edit: NOT SO
   c.  The total of all possible 6-subsets is C(15,6)

C(15,6)- C(10,6)=5050-210=4795

there is an error in my reasoning, b. is wrong, there are more 6-subsets than 210 , since they may contain a multiple of 3 ,like 1,2,4,5,7,9-  so I stop here,

but urge someone to evaluate the correct answer to b.




Edited on February 9, 2016, 10:16 pm
  Posted by Ady TZIDON on 2016-02-09 21:27:48

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