Answer to Question #142106 in Discrete Mathematics for Promise Omiponle

Question

12) Suppose that a room contains 10 cats and 15 dogs. How many ways are there to form a committee consisting of 6 of these animals if it must have more dogs than cats?

Answer 1

This committe must consist of 6 dogs or 5 dogs and 1 cat or 4 dogs and 2 cats. The number of ways of choosing $k$ elements from $n$ elements is equal to $\left(\begin{array}{c}n\\k\end{array}\right)=\frac{n!}{k!(n-k)!}$

Therefore, using combinatorial Sum and Product Rules we have that the number of ways to form a committee is

$\left(\begin{array}{c}15\\6\end{array}\right)+\left(\begin{array}{c}15\\5\end{array}\right)\left(\begin{array}{c}10\\1\end{array}\right)+\left(\begin{array}{c}15\\4\end{array}\right)\left(\begin{array}{c}10\\2\end{array}\right)=$

$=\frac{15!}{6!(9)!}+\frac{15!}{5!(10)!}10+\frac{15!}{4!(11)!}\frac{10!}{2!(8)!}=\frac{15\cdot14\cdot13\cdot12\cdot11\cdot10}{6\cdot5\cdot4\cdot3\cdot2}+ \frac{15\cdot14\cdot13\cdot12\cdot11}{5\cdot4\cdot3\cdot2}10+ \frac{15\cdot14\cdot13\cdot12}{4\cdot3\cdot2}\frac{10\cdot9}{2}=$

$=5,005+30,030+61,425=96,460$

Answer: $96,460$