Defective 35831

Among the 24 products, seven are defective. How many ways can we choose to check
a) 7 products so that they are all good
b) 7 products so that they are all defective
c) 3 good and two defective products?

Correct answer:

a =  19448
b =  1
c =  14280

Step-by-step explanation:

$$\begin{gathered} C_7(17)=\left(\genfrac{}{}{0ex}{}{17}{7}\right)=\frac{17!}{7!(17-7)!}=\frac{17\cdot 16\cdot 15\cdot 14\cdot 13\cdot 12\cdot 11}{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}=19448 \\ a=\left(\genfrac{}{}{0ex}{}{24-7}{7}\right)=19448 \end{gathered}$$

$$\begin{gathered} C_7(7)=\left(\genfrac{}{}{0ex}{}{7}{7}\right)=\frac{7!}{7!(7-7)!}=\frac{1}{1}=1 \\ b=\left(\genfrac{}{}{0ex}{}{7}{7}\right)=1 \end{gathered}$$

$$\begin{gathered} C_3(17)=\left(\genfrac{}{}{0ex}{}{17}{3}\right)=\frac{17!}{3!(17-3)!}=\frac{17\cdot 16\cdot 15}{3\cdot 2\cdot 1}=680 \\ {C}_2(7)=\left(\genfrac{}{}{0ex}{}{7}{2}\right)=\frac{7!}{2!(7-2)!}=\frac{7\cdot 6}{2\cdot 1}=21 \\ c=\left(\genfrac{}{}{0ex}{}{24-7}{3}\right)\cdot \left(\genfrac{}{}{0ex}{}{7}{2}\right)=\left(\genfrac{}{}{0ex}{}{24-7}{3}\right)\cdot 21=14280 \end{gathered}$$

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