Probability two blocks have exactly 2 out of 4 properties the same? 
I have 120 blocks. Each block is one of 2 different materials, 3 different colors, 4 different sizes, and 5 different shapes. No two blocks have exactly the same of all four properties. I take two blocks at random. What is the probability the two blocks have exactly two of these four properties the same?

I ended up getting:
$${{\binom{2}{1}\binom{3}{1}\binom{4}{2}\binom{5}{2} + \binom{2}{1}\binom{3}{2}\binom{4}{1}\binom{5}{2} + \binom{2}{1}\binom{3}{2}\binom{4}{2}\binom{5}{1} + \binom{2}{2}\binom{3}{1}\binom{4}{1}\binom{5}{2} + \binom{2}{2}\binom{3}{1}\binom{4}{2}\binom{5}{1} + \binom{2}{2}\binom{3}{2}\binom{4}{1}\binom{5}{1}}\over{\binom{120}{2}}} = {{41}\over{238}}$$Is this correct?
EDIT: The answer given here is ${{35}\over{119}} = {5\over{17}}$:
https://web2.0calc.com/questions/probability-question-blocks
Who is correct?
 A: You have made a simple arithmetic error. I evaluated your expression and I got $\dfrac{5}{17}$.
A: Yes.
Since no two blocks have all four properties identical, and $2\cdot 3\cdot 4\cdot 5=120$, there are exactly enough blocks for every combination of properties; thus each arrangement has identical probability.  Hence you can use this method.
The denominator counts the ways to select 2 among these 120 combinations, of course.  This is the size of the sample space.
Your numerator counts the ways to select one of two properties and two of the other for each of the six such arrangements of properties.  Thus properly counting the favoured outcomes.
So... everything checks out okay.
A: No. The answer given on the linked website is correct.
After picking the first block there are 119 left. Assume that the second block has material and color the same, whereas size and shape are different. Of the $4*5 = 20$ blocks that meet the first criterium (note: minus 1 because the first block can not be picked again!), it is easy to see that there are $(4-1)*(5-1) = 12$ that satisfy the second criterium. Analyzing the other combinations in the same way we get:
$$P = \frac {3*4 + 2*4 + 2*3 +1*4 + 1*3+1*2} {119} = \frac {12+8+6+4+3+2}{119} = \frac {35}{119}$$
