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.

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.

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*Edited on ***February 9, 2016, 10:16 pm**