Let C(n, k) = the number of k-membered subsets of an n-membered set. Find (a) C(6, k) for k = 0,1,2,...,6 (b) C(7, k) for k = 0,1.2...,7

Answer:(a) [tex]C(6,0) = 1[/tex], [tex]C(6,1) = 6[/tex], [tex]C(6,2) = 15[/tex], [tex]C(6,3) = 20[/tex], [tex]C(6,4) = 15[/tex], [tex]C(6,5) = 6[/tex], [tex]C(6,6) = 1[/tex].(b) [tex]C(7,0) = 1[/tex], [tex]C(7,1) = 7[/tex], [tex]C(7,2) = 21[/tex], [tex]C(7,3) = 35[/tex], [tex]C(7,4) = 35[/tex], [tex]C(7,5) = 21[/tex], [tex]C(7,6) = 7[/tex], [tex]C(7,7)=1[/tex].Step-by-step explanation:In this exercise we only need to recall the formula for C(n,k):[tex]C(n,k) = \frac{n!}{k!(n-k)!}[/tex]where the symbol [tex]n![/tex] is the factorial and means [tex]n! = 1\cdot 2\cdot 3\cdot 4\cdtos (n-1)\cdot n[/tex]. By convention 0!=1. The most important property of the factorial is [tex]n!=(n-1)!\cdot n[/tex], for example 3!=1*2*3=6.(a) The explanations to the solutions is just the calculations.[tex]C(6,0) = \frac{6!}{0!(6-0)!} = \frac{6!}{6!} = 1[/tex][tex]C(6,1) = \frac{6!}{1!(6-1)!} = \frac{6!}{5!} = \frac{5!\cdot 6}{5!} = 6[/tex][tex]C(6,2) = \frac{6!}{2!(6-2)!} = \frac{6!}{2\cdot 4!} = \frac{5!\cdot 6}{2\cdot 4!} = \frac{4!\cdot 5\cdot 6}{2\cdot 4!} = \frac{5\cdot 6}{2} = 15[/tex][tex]C(6,3) = \frac{6!}{3!(6-3)!} = \frac{6!}{3!\cdot 3!} = \frac{5!\cdot 6}{6\cdot 6} = \frac{5!}{6} = \frac{120}{6} = 20[/tex][tex]C(6,4) = \frac{6!}{4!(6-4)!} = \frac{6!}{4!\cdot 2!} = frac{5!\cdot 6}{2\cdot 4!} = \frac{4!\cdot 5\cdot 6}{2\cdot 4!} = \frac{5\cdot 6}{2} = 15[/tex][tex]C(6,5) = \frac{6!}{5!(6-5)!} = \frac{6!}{5!} = \frac{5!\cdot 6}{5!} = 6[/tex][tex]C(6,6) = \frac{6!}{6!(6-6)!} = \frac{6!}{6!} = 1[/tex].(b) The explanations to the solutions is just the calculations. [tex]C(7,0) = \frac{7!}{0!(7-0)!} = \frac{7!}{7!} = 1[/tex][tex]C(7,1) = \frac{7!}{1!(7-1)!} = \frac{7!}{6!} = \frac{6!\cdot 7}{6!} = 7[/tex][tex]C(7,2) = \frac{7!}{2!(7-2)!} = \frac{7!}{2\cdot 5!} = \frac{6!\cdot 7}{2\cdot 5!} = \frac{5!\cdot 6\cdot 7}{2\cdot 5!} = \frac{6\cdot 7}{2} = 21[/tex][tex]C(7,3) = \frac{7!}{3!(7-3)!} = \frac{7!}{3!\cdot 4!} = \frac{6!\cdot 7}{6\cdot 4!} = \frac{5!\cdot 6\cdot 7}{6\cdot 4!} = \frac{120\cdot 7}{24} = 35[/tex][tex]C(7,4) = \frac{7!}{4!(7-4)!} = \frac{6!\cdot 7}{4!\cdot 3!} = frac{5!\cdot 6\cdot 7}{4!\cdot 6} = \frac{120\cdot 7}{24} = 35[/tex][tex]C(7,5) = \frac{7!}{5!(7-2)!} = \frac{7!}{5!\cdot 2!} = 21[/tex][tex]C(7,6) = \frac{7!}{6!(7-6)!} = \frac{7!}{6!} = \frac{6!\cdot 7}{6!} = 7[/tex][tex]C(7,7) = \frac{7!}{7!(7-7)!} = \frac{7!}{7!} = 1[/tex]For all the calculations just recall that 4! =24 and 5!=120.