title: Combinations and probability
body: <p>So I'm having trouble with these two problems, and have kind of started them but not sure if I'm on the correct track. (I'm using the combination and permutation formulas)</p>
  
  <pre><code>Problem 1: "A group of students contains five men and six women."
  "I go to the store to buy hats for the eleven students. Hats come in four colors: red, blue,
  green, and yellow. How many different ways can I buy the hats, if I have to buy at least one hat
  each color?" 
  </code></pre>
  
  <p>This is how I'm going about the problem:</p>
  
  <p>Total ways buy hats with 1 of each color = Total ways of buying hats - Total ways of buying only 3 different kind of hats - Total ways of buying only 2 different kind of hats - Total ways of buying only 1 kind of hat</p>
  
  <p>Total ways of buying hats repetition C(4, 11): <strong>!(4 + 11 - 1)/!11(4-1)!</strong> </p>
  
  <p>Total ways of buying 3 hats: 11-combinations of buying 3 hats, C(3, 11),  * 3-combinations of 4 hats, C(4,3):
  <strong>(!(3 + 11 - 1)/!11(3-1)!) * (!4/!3(4-3)!)</strong></p>
  
  <p>.... and so on for 2 hats and 1 hats. </p>
  
  <p>Is this correct? Thinking about it logically, shouldn't the difference of that be the answer?</p>
  
  <p>Sorry if that's hard to understand my work, it was the best I could put it in normal format. But pretty much what I'm doing is getting the total choices by repetitive C(4, 11), and then subtracting the total choices of 3, 2, and 1 hats chosen.<br>
  To find those total choices I do the<br>
  the repetitive C(X,11) * non-repetitive C(4, X) for x-values 3,2,1</p>
  
  <p>-thanks</p>