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.

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)

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

edited: 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