Equations involving indices of binomial coefficients

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There's always brute force (there is only a small finite number of possibilities for $y$):

sage: var('z')
sage: rhs = sum(z*binomial((1001-z),950),z,1,51)
sage: next(y for y in range(1, 1002) if binomial(1002, y) == rhs)
50
sage: [y for y in range(1, 1002) if binomial(1002, y) == rhs]
[50, 952]

"Brute-force" solution :

sage: var("y, z")
(y, z)

The right-hand size involves a constant ; compute it once...

sage: R = sum(z*binomial((1001-z),950),z,1,51) ; R
10480853106371895870377052872259391385423261386799458513700836377945732866106336546780

Brute force check of all possible (integer) solution candidates :

sage: [u for u in range(1003) if binomial(1002,u)==R]
[50, 952]

If this is homework, your professor may or may not appreciate this answer : there might be subtler points in his/her course...

HTH nevertheless...