## Precalculus (10th Edition)

a) $3003$ b) $20475$ c)$16653$
If we have two groups with $n$ and $m$ objects and we choose $r(r\leq n)$ and $s(s\leq m)$ objects from the two groups where the order doesn't matter then the number of combinations is: $C(n,r)C(m,s).$ (Similarly for more than $2$ groups.) We know that $C(n,r)=\frac{n(n-1)(n-2)...(n-k+1)}{r!}$. Also $C(n,0)=1$ by convention. Also $C(n,r)=C(n,n-r).$ a)Hence here $C(15,5)C(10,0)=\frac{15\cdot14\cdot13\cdot12\cdot11}{5\cdot4\cdot3\cdot2\cdot1}\cdot1=3003$. b)Hence here $C(15,3)C(10,2)=\frac{15\cdot14\cdot13}{3\cdot2\cdot1}\cdot\frac{10\cdot9}{2\cdot1}=455\cdot45=20475$. c)Hence here $C(15,4)C(10,1)=\frac{15\cdot14\cdot13\cdot12}{4\cdot3\cdot2\cdot1}\cdot\frac{10}{1}=1365\cdot10=13650$ but we must add the result from a) too, hence the total: $13650+3003=16653$