# Divide

How many different ways can three people divide 7 pears and 5 apples?

Result

x =  2940

#### Solution:

$C_{{ 3}}(9) = \dbinom{ 9}{ 3} = \dfrac{ 9! }{ 3!(9-3)!} = \dfrac{ 9 \cdot 8 \cdot 7 } { 3 \cdot 2 \cdot 1 } = 84 \ \\ n_{1}=7 \ \\ n_{2}=5 \ \\ k=3 \ \\ C_{{ 3}}(7)=\dbinom{ 7}{ 3}=\dfrac{ 7! }{ 3!(7-3)!}=\dfrac{ 7 \cdot 6 \cdot 5 } { 3 \cdot 2 \cdot 1 }=35 \ \\ \ \\ x={ { n_{1}+k-1 } \choose k } \cdot \ { { n_{2}+k-1 } \choose k }={ { 7+3-1 } \choose 3 } \cdot \ { { 5+3-1 } \choose 3 }=2940$

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Showing 1 comment:
Math student
I think that you mixed up n and k. My understanding is that : n = 3, k1 = 7, k2 = 5

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