There was some understandable confusion as to what the 199'th row of Pascal's triangle is.  Most working mathematicians agree that the n'th row of Pascal's triangle starts with the entries 1, n ,...; the row containing just a 1 is generally thought of as the zero'th row.

Again, this is the kind of problem which is easier to solve in for an arbitrary row than in the specific case of the 199th row.  The entries in the n'th row Pascal's triangle are the coefficients of the polynomial (1 + x)ⁿ.  From the binomial theorem, we have .  Therefore, determining the number of odd entries in the n'th row of Pascal's triangle is the same a determining the number of odd coefficients in the polynomial (1 + x)ⁿ which is further the same as determining the number of non-zero coefficients in (1 + x)ⁿ (modulo 2).  

An easy induction shows that (1 + x)²ⁿ = (1 + x²ⁿ) (modulo 2).  A deep theorem in mathematics gives that 199 = 128 + 64 + 4 + 2 + 1.  Putting there facts together gives that (1 + x)¹⁹⁹ = (1 + x)¹²⁸(1 + x)⁶⁴(1 + x)⁴(1 + x)²(1 + x) = (1 + x¹²⁸)(1 + x⁶⁴)(1 + x⁴)(1 + x²)(1 + x) (modulo 2) which is a polynomial with 2⁵ = 32 non-zero terms.  The general result is that the number of odd entries in the n'th row of Pascal's triangle is 2 to the number of ones in the binary expansion of n.

I recommend the applet at the website www.math.ohio-state.edu/~btk/Pascal/ for those wanting to see what geometric questions this and related problems create.

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ã2004 Alberto L. Delgado