The number of ways 16 identical cubes,of whic | vedclass

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(A) We have $5$ red cubes and $11$ blue cubes. Let the red cubes be $R$. Placing $5$ red cubes in a row creates $6$ gaps (including the ends) where bl

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$56$

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(A) We have $5$ red cubes and $11$ blue cubes. Let the red cubes be $R$. Placing $5$ red cubes in a row creates $6$ gaps (including the ends) where blue cubes can be placed: $\_ R \_ R \_ R \_ R \_ R \_$.Let $x_1, x_2, x_3, x_4, x_5, x_6$ be the number of blue cubes in these $6$ gaps.We have the equation $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 11$.The condition that there must be at least $2$ blue cubes between any two red cubes implies $x_2, x_3, x_4, x_5 \geq 2$,while $x_1, x_6 \geq 0$.Let $x_2 = t_2 + 2, x_3 = t_3 + 2, x_4 = t_4 + 2, x_5 = t_5 + 2$,where $t_2, t_3, t_4, t_5 \geq 0$.Substituting these into the equation: $x_1 + (t_2 + 2) + (t_3 + 2) + (t_4 + 2) + (t_5 + 2) + x_6 = 11$.$x_1 + t_2 + t_3 + t_4 + t_5 + x_6 + 8 = 11$.$x_1 + t_2 + t_3 + t_4 + t_5 + x_6 = 3$.Using the stars and bars formula,the number of non-negative integer solutions is given by $\binom{n+k-1}{k-1}$,where $n=3$ and $k=6$.Number of ways $= \binom{3+6-1}{6-1} = \binom{8}{5} = \binom{8}{3} = \frac{8 	imes 7 	imes 6}{3 	imes 2 	imes 1} = 56$.

The number of ways $16$ identical cubes,of which $11$ are blue and the 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

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