A bakery currently has 5 pies and 6 cakes in its inventory. The bakery’s owner has decided to display 5 of these items in the bakery’s front window. If the items are randomly selected, what is the probability that the display will have exactly 3 pies?  Select the correct option:
A) 3/11 B) 25/77 C) 5/11 D) 7/11 E) 50/77
We need the probability that the owner will select PPPCC: 5!/3!∗2!∗(5/11∗4/10∗3/9)∗(6/8∗5/7)=25/77, we are multiplying by 5!/3!∗2! since PPPCC scenario can occur in number of ways: PPPCC, PPCPC, PCPPC, CPPPC, ... Notice that the number of occurrences of PPPCC basically is the number of arrangements of 5 letters PPPCC out of which 3 P's and 2 C's are identical, so it's 5!/3!∗2!   Answer: 25/77.
<result>B</result>
A bakery currently has 5 pies and 6 cakes in its inventory. The bakery’s owner has decided to display 5 of these items in the bakery’s front window. If the items are randomly selected, what is the probability that the display will have exactly 3 pies?