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When question is "What is the number of ways to create a group of 5 from 17 girls and 20 boys when you have to select at least 3 girls

I needs a simple explanation why it is that $$ {17\choose3} {20\choose2} + {17\choose 4} {20 \choose 1} + {17 \choose 5} $$ (3 girls with 2 boys ... All girls) Which is the correct answer and I completely understand the logic

VS $$ {17 \choose 3} { 34 \choose 2} $$ (3 girls and random 2 from the remaining 34) Which is incorrect but I don't understand why it give a different number that the first

Matti P.
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Shaun Vermaak
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    The second one is incorrect because it over-counts: If you picked a,b,c first and then d,e, that gives you the same set as picking a,b,d and then c,e etc.. – Aravind May 11 '20 at 10:26
  • And the over counting is not uniform: GGGBB combos are ok in your second method, but GGGGB's are over-counted by a factor of $4$, and GGGGG's by a factor of $10$ -- so fixing up the over-count means splitting up the types as done the first solution, anyway. – Ned May 11 '20 at 11:38

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