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Find the coefficient of $a^5b^5c^5d^6$ in the expansion $(bcd+acd+abd+abc)^7$.

I tried to use multinomial theorem but failed.I can find out the coefficient of $a^5b^5c^5d^6$ in $(a+b+c+d)^21$.But I discovered that the multiplication of terms in the given expansion is $(abcd)^21$ and in my expression I have done the same operation and have found the same result.Is there any relation between these two expressions??I am sure that it can be done using multinomial theorem but how??If you have another method then show me that also.

*The 2nd expression which I gave has a big answer.The answer is so bigger than the original question.Please help me.Thanks

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Notice that if we have:

$$(abc)^i(abd)^j(bcd)^k(acd)^l=a^5b^5c^5d^6 ;$$

then this implies that

$$ \left\{ \begin{array}{lcc} i+j+l=5 , & \text{by comparison of the power of} \ \ a , \\ i+j+k=5 , & \text{by comparison of the power of} \ \ b , \\ i+k+l=5 , & \text{by comparison of the power of} \ \ c , \\ j+k+l=6 , & \text{by comparison of the power of} \ \ d . \\ \end{array} \right.$$


The above system of equations has the solution:

$$i=1, \ \ j=2, \ \ k=2, \ \ l=2.$$


So we have the following:

$$(abc)^1(abd)^2(bcd)^2(acd)^2=a^5b^5c^5d^6.$$


So the coefficeint is equal to $\dfrac{7!}{1!2!2!2!}$.

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    $\begingroup$ it is okay bro.But why u distribute $a^5b^5c^5d^6$ into$(abc)^1(abd)^2(bcd)^2(acd)^2?It can be distributed in another manner too!But it is so simple.I actually failed to distribute $a^5b^5c^5d^6$ in that manner.Thanks $\endgroup$ – Sufaid Saleel Aug 17 '17 at 16:00
  • $\begingroup$ @Sufaid Saleel , Now I have explained more. Does it satisfies you? $\endgroup$ – Davood Khajehpour Aug 17 '17 at 16:13
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    $\begingroup$ Thank you .The total sum is completely cleared.Thanks for your cooperation $\endgroup$ – Sufaid Saleel Aug 17 '17 at 16:31
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The coefficient of $a^5b^5c^5d^6$ in $(bcd+acd+abd+abc)^7$ is the same as the coefficient of $\dfrac{a^5b^5c^5d^6}{(abcd)^7}$ in $\dfrac{(bcd+acd+abd+abc)^7}{(abcd)^7}$, i.e. the coefficient of $\dfrac1{a^2b^2c^2d}$ in $\left(\dfrac1a+\dfrac1b+\dfrac1c+\dfrac1d\right)^7$. Is that better for you?

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  • $\begingroup$ It is Enough for me.I ask it for increase my own knowledge.But is it a good question??What do u think?? $\endgroup$ – Sufaid Saleel Aug 17 '17 at 15:54

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