A bakery currently has 5 pies and 6 cakes in its inventory

Combination approach:

\(\frac{C^3_5*C^2_6}{C^5_{11}}=\frac{25}{77}\), where:
\(C^3_5\) is # of ways to chose 3 pies out of 5;
\(C^2_6\) is # of ways to chose 2 cakes out of 6;
\(C^5_{11}\) is # of ways to chose 5 items out of total 5+6=11.

Answer: B.

Probability approach:

We need the probability that the owner will select PPPCC: \(\frac{5!}{3!*2!}*(\frac{5}{11}*\frac{4}{10}*\frac{3}{9})*(\frac{6}{8}*\frac{5}{7})=\frac{25}{77}\), we are multiplying by \(\frac{5!}{3!*2!}\) 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 \(\frac{5!}{3!*2!}\) .

Answer: B.

Hope it's clear.