Four counting problems

Problem

a) A lady buys $12$ bottles choosing among $4$ types of wine. In how many ways can she fill her cart? b) A coach has $19$ players and must field $11$ starters, assigning each a specific position. How many different line-ups? c) In how many ways can you split $24$ million (in whole millions) across $5$ different accounts? d) From $10$ sons and $12$ daughters, $5$ and $4$ are chosen respectively. In how many ways?

Solution

a) Combinations with repetition of $4$ types in class $12$: $\dbinom{12+4-1}{4-1}=\dbinom{15}{3}=455$. b) Simple arrangements (distinct positions, no repetition): $D_{19,11}=\dfrac{19!}{8!}=3\,016\,991\,577\,600$. c) Splitting $24$ units into $5$ integer addends $\ge0$: $\dbinom{24+5-1}{5-1}=\dbinom{28}{4}=20475$. d) Product of two combinations: $\dbinom{10}{5}\dbinom{12}{4}=252\cdot495=124740$. $\boxed{\,a)\ 455\,;\quad b)\ 3\,016\,991\,577\,600\,;\quad c)\ 20475\,;\quad d)\ 124740\,}$