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  • The number of ways, 16 identical cubes, of which 11 are blue and rest are red, can be placed in a row so that between any two red cubes there should be at least 2 blue cubes, is_____________.<d

The number of ways, 16 identical cubes, of which 11 are blue and rest are red, can be placed in a row so that between any two red cubes there should be at least 2 blue cubes, is_____________.

Option: 1

28


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Let  \mathrm{n_{1},n_{2},n_{3},n_{4},n_{5}\: and\: n_{6}}  are number of blue balls in the gaps of red balls

Here \mathrm{n_{1} \geqslant 0, n_{2} \geqslant 2, n_{3} \geqslant 2, n_{4} \geqslant 2, n 5 \geqslant 2, n_{6} \geqslant 0}

Take \mathrm{ n_{1}=n_{1}^{1}, n_{2}-2=n_{2}^{1}, n_{3}-2=n_{3}^{1}, n_{4}-2-n_{4}^{1} , n_{5}-2=n_{5}^{1}, n_{6}=n_{6}^{1} }

\mathrm{n_{1}+n_{2}+n_{3}+n_{4}+n_{5}+n_{6}=11} \\

\mathrm{\Rightarrow n_{1}+\left(n_{2}-2\right)+\left(n_{3}-2\right)+\left(n_{4}-2\right)+\left(n_{5}-2\right)+n_{6}=11-8} \\

\mathrm{\Rightarrow n_{1}{ }^{\prime}+n_{2}^{\prime}+n_{3}{ }^{\prime}+n_{4}{ }^{\prime}+n_{5}{ }^{\prime}+n_{6}{ }^{\prime}=3 }\\

no of non-negative integral solutions

\mathrm{=^{6+3-1} C_{3-1}={ }^{8} C_{2}=28}

Hence answer is \mathrm{28}

 

Posted by

Ajit Kumar Dubey

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