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Question: What will be the coefficients of \({a^8}{b^4}{c^9}{d^9}\) in \({\left( {abc + abd + acd + bcd} \righ...

What will be the coefficients of a8b4c9d9{a^8}{b^4}{c^9}{d^9} in (abc+abd+acd+bcd)10{\left( {abc + abd + acd + bcd} \right)^{10}}.

Explanation

Solution

Hint: In this question use the direct formula for any general term for equation in form of (abc+abd+acd+bcd)n{\left( {abc + abd + acd + bcd} \right)^n}which is n!x!.y!.z!.q!(abc)x(abd)y(acd)z(bcd)q=n!x!.y!.z!.q!a(x+y+z)b(x+y+q)c(x+z+q)d(y+z+q)\dfrac{{n!}}{{x!.y!.z!.q!}}{\left( {abc} \right)^x}{\left( {abd} \right)^y}{\left( {acd} \right)^z}{\left( {bcd} \right)^q} = \dfrac{{n!}}{{x!.y!.z!.q!}}{a^{\left( {x + y + z} \right)}}{b^{\left( {x + y + q} \right)}}{c^{\left( {x + z + q} \right)}}{d^{\left( {y + z + q} \right)}}.The direct power and coefficients comparison will get to the answer.

Complete step-by-step answer:

As we know the general term of (abc+abd+acd+bcd)10{\left( {abc + abd + acd + bcd} \right)^{10}} is
10!x!.y!.z!.q!(abc)x(abd)y(acd)z(bcd)q=10!x!.y!.z!.q!a(x+y+z)b(x+y+q)c(x+z+q)d(y+z+q)\dfrac{{10!}}{{x!.y!.z!.q!}}{\left( {abc} \right)^x}{\left( {abd} \right)^y}{\left( {acd} \right)^z}{\left( {bcd} \right)^q} = \dfrac{{10!}}{{x!.y!.z!.q!}}{a^{\left( {x + y + z} \right)}}{b^{\left( {x + y + q} \right)}}{c^{\left( {x + z + q} \right)}}{d^{\left( {y + z + q} \right)}}
Now we need the coefficient of a8b4c9d9{a^8}{b^4}{c^9}{d^9}
So on comparing this with above equation we have,
x+y+z=8\Rightarrow x + y + z = 8................... (1)
x+y+q=4\Rightarrow x + y + q = 4................... (2)
x+z+q=9\Rightarrow x + z + q = 9................... (3)
y+z+q=9\Rightarrow y + z + q = 9.................. (4)
Now add all the four equation we have,
3x+3y+3z+3q=8+4+9+9=30\Rightarrow 3x + 3y + 3z + 3q = 8 + 4 + 9 + 9 = 30
Now divide by 3 we have,
x+y+z+q=10\Rightarrow x + y + z + q = 10 ............................. (5)
Now subtract equation (1), (2), (3) and (4) from equation (5) respectively we have,
x+y+z+qxyz=108\Rightarrow x + y + z + q - x - y - z = 10 - 8
q=2\Rightarrow q = 2
And
x+y+z+qxyq=104\Rightarrow x + y + z + q - x - y - q = 10 - 4
z=6\Rightarrow z = 6
And
x+y+z+qxzq=109\Rightarrow x + y + z + q - x - z - q = 10 - 9
y=1\Rightarrow y = 1
And
x+y+z+qyzq=109\Rightarrow x + y + z + q - y - z - q = 10 - 9
x=1\Rightarrow x = 1
Therefore the coefficient of a8b4c9d9{a^8}{b^4}{c^9}{d^9} is
10!1!.1!.6!.2!a8b4c9d9\Rightarrow \dfrac{{10!}}{{1!.1!.6!.2!}}{a^8}{b^4}{c^9}{d^9}
Now simplify the above equation we have,
10×9×8×7×6!1×1×6!×2×1a8b4c9d9\Rightarrow \dfrac{{10 \times 9 \times 8 \times 7 \times 6!}}{{1 \times 1 \times 6! \times 2 \times 1}}{a^8}{b^4}{c^9}{d^9}
10×9×8×72a8b4c9d9\Rightarrow \dfrac{{10 \times 9 \times 8 \times 7}}{2}{a^8}{b^4}{c^9}{d^9}
2520a8b4c9d9\Rightarrow 2520{a^8}{b^4}{c^9}{d^9}
So the required coefficient of a8b4c9d9{a^8}{b^4}{c^9}{d^9} in (abc+abd+acd+bcd)10{\left( {abc + abd + acd + bcd} \right)^{10}} is 2520.
So this is the required answer.

Note: Such types of questions are direct formula based and it is always advised to remember these direct formulas. It’s not a binomial expansion so we need not to be confused between these two concepts. Both are different to each other, any general term in the binomial expansion of (x+y)n{\left( {x + y} \right)^n}is its (nr+2)th{\left( {n - r + 2} \right)^{th}} term.