In the shipment

There are 40 products in the shipment. Of these, 4 are defective. In how many ways can we select 5 products so that exactly 3 of them are good?

Correct answer:

n =  42840

Step-by-step explanation:

$$\begin{gathered} C_3(36)=\left(\genfrac{}{}{0ex}{}{36}{3}\right)=\frac{36!}{3!(36-3)!}=\frac{36\cdot 35\cdot 34}{3\cdot 2\cdot 1}=7140 \\ {n}_1=\left(\genfrac{}{}{0ex}{}{40-4}{3}\right)=7140 \\ {C}_2(4)=\left(\genfrac{}{}{0ex}{}{4}{2}\right)=\frac{4!}{2!(4-2)!}=\frac{4\cdot 3}{2\cdot 1}=6 \\ {n}_2=\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6 \\ n=n_1\cdot n_2=7140\cdot 6=42840 \end{gathered}$$

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