# First class

The shipment contains 40 items. 36 are first grade, 4 are defective. How many ways can select 5 items, so that it is no more than one defective?

Correct result:

n =  612612

#### Solution:

$C_{{ 5}}(36)=\dbinom{ 36}{ 5}=\dfrac{ 36! }{ 5!(36-5)!}=\dfrac{ 36 \cdot 35 \cdot 34 \cdot 33 \cdot 32 } { 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 }=376992 \ \\ \ \\ n_{1}={ { 36 } \choose 5 }=376992 \ \\ C_{{ 4}}(36)=\dbinom{ 36}{ 4}=\dfrac{ 36! }{ 4!(36-4)!}=\dfrac{ 36 \cdot 35 \cdot 34 \cdot 33 } { 4 \cdot 3 \cdot 2 \cdot 1 }=58905 \ \\ \ \\ n_{2}=4 \cdot \ { { 36 } \choose 4 }=4 \cdot \ 58905=235620 \ \\ n=n_{1}+n_{2}=376992+235620=612612$

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