(a)\({6 \choose 3}\) = 20
choose(6,3)## [1] 20
\({n \choose j}=(n)_j/j!=n!/j!(n-j)!\) = 6!/3!(6-3)!= 20
factorial(6)/(factorial(3)*factorial(6-3))## [1] 20
\(b(n, p, j) = {n \choose j}p^jq^nāj\) where \(q = (1-p)\)
= \({5 \choose 4}(1/5)^4*(4/5)^1\)
choose(5,4)*(1/5)**4*(4/5)## [1] 0.0064
(c)\({7 \choose 2}\) = 35
(d)\({26 \choose 26}\) = 1
choose(4,3)*(1/5)**3*(4/5)## [1] 0.0256
(f)\({6 \choose 2}\) = 15
(g)\({10 \choose 9}\) = 10
choose(8,5)*(.3)**5*(.7)**3## [1] 0.04667544