Strangely deep question (1 Viewer)

[no_arg]

Find all n and k such that (n choose k) is the sum of (13 choose 6) and (13 choose 5).

 

[hellohi786364]

13C6 + 13C5 = 13!/6!7! + 13!/5!8!
= (8(13!) + 6(13!))/8!6!

=14!/8!6!
=14!/8!(14-8)! or 14!/6!(14-6)!
=14C8 or 14C6

or

=15!/8!6!(5x3)
=15!/5!(3x3x2x5x8!)
=15!/5!(8!x9x10)
=15!/5!10!
=15C10 or 15C5

or

=3003
=3003C1 or 3003C3002

or

=3003
=3x7x11x13
=((7x11)x(13x3)x2)/2
=(77x78)/2!
=78!/2!76!
=78C2 or 78C76

 

[no_arg]

13C6 + 13C5 = 13!/6!7! + 13!/5!8!
= (8(13!) + 6(13!))/8!6!

=14!/8!6!
=14!/8!(14-8)! or 14!/6!(14-6)!
=14C8 or 14C6

or

=15!/8!6!(5x3)
=15!/5!(3x3x2x5x8!)
=15!/5!(8!x9x10)
=15!/5!10!
=15C10 or 15C5

or

=3003C1 or 3003C3002

or

=78C2 or 78C76

Yes, these sorts of questions often appear in trial papers as if they were trivial and unique, when in fact they are generally only solvable by exhaustion with multiple solutions. Could there ever be more than 8 solutions? See Singmaster Conjecture.