I'm afraid my argument is a bit involved, and maybe someone else can post a simpler explanation.

Without loss of generality, let the non-white objects be labeled as black. So we have a multiset (hopefully there's no confusion that I re-use the letter W):

S = {W,...,W,B,...,B}

For example, if M = 7 and W = 4 then we have

S = {W, W, W, W, B, B, B}

Just say we pick them all out, without replacement. Then any particular order is given by a permutation of S. You could think in terms of 7-tuples or strings; for example, one possible permutation is

(W, W, B, B, W, B, W) (tuple representation)

WWBBWBW (string representation)

Now I claim that

1) The probability that the ith index is W in the above example is the same as the probability that the ith index is W if we had only picked n elements, instead of all of them.

2) The probability that the ith index is W in the above example is independent of i (that is, the same no matter what i is).

I won't give formal proofs, but more hand-wavy explanations

1) This is true because the probability of an event is not dependent on events that happen afterwards.

2) This is true because there is nothing special to distinguish any index from any other index. So consider permutations of the set {1,2,3,4,5}. It is intuitively clear that the number 1 is equally likely to appear in any index in the range {1...5} for a randomly chosen permutation. If we found that 1 is more likely to appear in the 2nd position compared with the 4th, we would naturally suspect we had made an error in counting.

There's probably a much simpler explanation, and the lack of formal proof I offered may leave some unconvinced, nevertheless I believe it to be valid reasoning.

(One final note: We are assuming that i <= n. Obviously if i > n then the probability could either be considered 0, or undefined since the event is not in the sample space.)