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in Combinatory recategorized by
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$24?$
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let $x_1$ be the number of green balls

let $x_2$ be the number of yellow balls

let $x_3$ be the number of blue balls

$x_1 + x_2 + x_3 = 10                       ,$         $0 \leq x_1<11, \ \ 0 \leq x_2<6, \ \ \ 0 \leq x_3<4$

Total possible solutions without any restriction = $\binom{12}{2} = 66$

Now let's calculate number of invalid solutions,

Total solutions when $\color{red}{x_2 \geq 6,}$

$x_1 + x_2 + 6 + x_3 = 10$

 $x_1 + x_2 + x_3 = 4$

$ = \binom{6}{2} = 15$

 

Total solutions when $\color{red}{x_3 \geq 4,}$

$x_1 + x_2 + x_3 + 4 = 10$

 $x_1 + x_2 + x_3 = 6$

$ = \binom{8}{2} = 28$

 

Total solutions when both $\color{red}{x_3 \geq 4}$ and $\color{red}{x_2 \geq 6,}$

$x_1 + x_2+  6 + x_3 + 4 = 10$

$x_1 + x_2 + x_3 = 0$

$ = \binom{2}{2} = 1$

 

Total invalid solutions $= 15 + 28 - 1 = 42$

Total valid solutions = $66 - \color{red}{42} = 24$
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3 Comments

why you did not exclude x1>=11
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 because that's not a possibility, total balls to be chosen are just 10

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ohhhh!! thank you so much
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