1. ## Arbitary Arrangement

I need help again

Question is as follows:

Twelve basketball players whose uniforms are numbered 1 through 12 stand in a line in an arbitrary arrangement. prove that there must be three consecutive players in the line such that the sum of the numbers of their uniforms is at least 20.

2. Originally Posted by faisalnet5
I need help again

Question is as follows:

Twelve basketball players whose uniforms are numbered 1 through 12 stand in a line in an arbitrary arrangement. prove that there must be three consecutive players in the line such that the sum of the numbers of their uniforms is at least 20.

$\displaystyle 1+2+..+12=78$
considering the (1,2,3) (4,5,6)(7,8,9)(10,11,12),four 3-tuples.((1,2,3)means the consecutive series formed by the first,second,third people)
since $\displaystyle 78\div 4=19.5$ and the sum of the numbers in (1,2,3),(4,5,6),(7,8,9),(10,11,12) is 78, so one one them must be greater than 19.5, so greater than 20.

3. But Yuj....how did u get 156\div 4=19.5?

Regards

5. Originally Posted by faisalnet5

Regards
there is a mistake....$\displaystyle 1+2+..+12=78$...sorry

6. Hello faisalnet5
Originally Posted by faisalnet5