## College Algebra (10th Edition)

$1; n$
RECALL: ${n} \choose {k}$ $= \dfrac{n!}{k!(n-k)!}$ Use the formula above to obtain: ${n} \choose {0}$ $\require{cancel}=\dfrac{n!}{0!(n-0)!} = \dfrac{n!}{1(n!)}=\dfrac{\cancel{n!}}{1(\cancel{n!})}=1$ ${n} \choose {1}$ $\require{cancel}=\dfrac{n!}{1!(n-1)!}=\dfrac{n!}{(n-1)!}=\dfrac{n\cancel{(n-1)(n-2)(n-3)....(3)(2)(1)}}{\cancel{(n-1)(n-2)(n-3)....(3)(2)(1)}}=n$ Thus, the missing expressions in the given statement (in order) are: $1; n$