<s>[INST] Form groups of people using combinatorics I have this problem: 

How many groups of 5 people can be formed between 4 boys and 7 girls
  if there should be at least 2 girls included?

My attempt was:
Ways of select two girls of a total of $7$ is: $\binom{7}{2} = 21$
Ways to select the $3$ remaining people, out of a total of $4$ boys and $5$ girls, is: $\binom{9}{3} = 84$
Therefore, $21 \cdot 84 = 1764$, but according to the guide my answer is incorrect. 
What is wrong with my development? Thanks in advance. 
 [/INST] There are four cases, with 2, 3, 4, or 5 girls in the 5-group.
The solution is: 
$$ {4 \choose 3}{7 \choose 2} + 
  {4 \choose 2}{7 \choose 3} + 
   {4 \choose 1}{7 \choose 4} + 
  {4 \choose 0}{7 \choose 5} = 455 
$$
What you have counted is :
$$ {4 \choose 3}{7 \choose 2} {\color{red} {2 \choose 2}} + 
   {4 \choose 2}{7 \choose 3} {\color{red} {3 \choose 2}} + 
   {4 \choose 1}{7 \choose 4} {\color{red} {4 \choose 2}} + 
   {4 \choose 0}{7 \choose 5} {\color{red} {5 \choose 2}} = 1764 
$$
meaning that between the choosen girls there are two special, i.e. the two ones selected in the beggining to ensure the quota.
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