ધારો કે A =left[ a _{i j}right], a _{i j} in | vedclass

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Question

As given $a+b+c+d=3$ or $5$ or $7$ or $11$

$	ext { if sum }=3$

$\left(1+x+x^2+\ldots++x^4ight)^4 ightarrow x^3$

$\left(1-x^5ight)^4(1-

Vedclass · Accepted Answer

$204$

Vedclass · Answer

As given $a+b+c+d=3$ or $5$ or $7$ or $11$

$	ext { if sum }=3$

$\left(1+x+x^2+\ldots++x^4ight)^4 ightarrow x^3$

$\left(1-x^5ight)^4(1-x)^{-4} ightarrow x^3$

$	herefore{ }^{4+3-1} C_3={ }^6 C_3=20$

If $\operatorname{sum}=5$

$\left(1-4 x^5ight)(1-x)^{-4} ightarrow x^5$

$\Rightarrow{ }^{4+5-1} C_5-4 x^{4.4+0-1} C_0={ }^8 C_5-4=52$

If sum $=7$

$\left(1-4 x ^5ight)(1- x )^{-4} ightarrow x ^7$

$\Rightarrow{ }^{4+5-1} C _4-{ }^{4 \cdot 4+0-1} C _0={ }^8 C _5-4=52$

$	ext { If sum }=11$

$\quad\left(1-4 x ^5+6 x ^{10}ight)(1- x )^{-4} ightarrow x ^{11}$

$\Rightarrow{ }^{4+11-1} C _{11}-4 \cdot{ }^{4+6-4} C _6+6 \cdot{ }^{4+1-1} C _1$

$={ }^{14} C _{11}-4 \cdot{ }^9 C _6+6.4=364-336+24=52$

$	herefore 	ext { Total matrices }=20+52+80+52=204$

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