Consider three positive integers x, y, and z that satisfy this equation:
comb(x,y)*comb(y,z) = 2*comb(x,z)
Does the above equation possess a finite number of solutions?
If so, determine all possible solutions of the above mentioned equation.
If not - then derive, with proof, the possible relationship between x, y, and z.
Note: comb(m,n) is the number of ways of choosing n unordered outcomes from m possibilities and defined as:
m!
comb(m,n) = ---------
n!(m-n)!
It is also known as binomial coefficient and read as "m choose n".
My proof is essentially the same as Brian's but I want to share some thoughts.
The choice of
2 in the equation is interesting as it's the only number that appears only once in Pascal's Triangle and yields the triple (x,x-1,x-2)
Replace it with 6, which appears three times C(4,2)=C(6,1)=C(6,5)
and you get three infinite outcomes for (x,y,z):
(x,x-2,x-4), (x,x-1,x-6), and (x,x-5,x-6)
Another implication is that for any choice of (x,y,z) the product C(x,y)*C(y,z) is a multiple of C(z,y). Specifically it's C(x-z,x-y)
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Posted by Jer
on 2022-11-08 08:27:20 |
further thoughts
