
Find the coefficient of ${a^2}{b^3}{c^4}d$ in the expansion of ${\left( {a - b - c + d} \right)^{10}}$.
Answer
474.9k+ views
Hint: Start by grouping the variables inside the expression , then try to find out the highest value of r to be used while expanding the expression . Use the same procedure for further expansion in order to get the desired powers of a, b ,c and d . Separate the coefficient along with the sign obtained and solve it in order to find the required value.
Complete step-by-step answer:
Given,
${\left( {a - b - c + d} \right)^{10}}$
Let us assume x = (a – b) and y = (d – c)
So the given equation would become as below,
${(x + y)^{10}}$
Now, That we need to find the coefficients of ${a^2}{b^3}{c^4}d$
We know , sum of powers or superscripts of a, b, c and d is 5.
Which means we need to check for those coefficients only which make the sum of powers of a, b, c, d as 5.
So we will proceed further by using the binomial expansion formula for ${(p + q)^n}$
${}^n{C_r}{p^r}{q^{n - r}}$where ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
So, for ${(x + y)^{10}}$=
${}^{10}{C_5}{x^5}{y^5}$
Substituting the values of x and y, we get
$ \Rightarrow {}^{10}{C_5}{(a - b)^5}{(d - c)^5}$
Now applying binomial expansion for both the terms again as per the required exponents of a, b , c, d ,we get
$ \Rightarrow {}^{10}{C_5} \cdot \left[ {{}^5{C_2} \cdot {a^2} \cdot {{( - b)}^3}} \right] \cdot \left[ {{}^5{C_1}{{( - c)}^4} \cdot d} \right]$
Rearranging the terms we get
$\Rightarrow - {}^{10}{C_5} \cdot {}^5{C_2} \cdot {}^5{C_1} \cdot {a^2} \cdot {b^3} \cdot {c^4} \cdot d$
Now , the coefficients of ${a^2}{b^3}{c^4}d$ will be,
$
= - {}^{10}{C_5}.{}^5{C_2}.{}^5{C_1} \\
= - 12600 \\
$
So , the coefficient of ${a^2}{b^3}{c^4}d$ is\[ - 12600\].
Note: Such similar questions can be solved by the same approach used as above. Attention must be given while expanding the binomial expression as any wrong input might lead to vague or incorrect answers. Also ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ value can never be negative, but the coefficients can be due to the nature of variable.
Complete step-by-step answer:
Given,
${\left( {a - b - c + d} \right)^{10}}$
Let us assume x = (a – b) and y = (d – c)
So the given equation would become as below,
${(x + y)^{10}}$
Now, That we need to find the coefficients of ${a^2}{b^3}{c^4}d$
We know , sum of powers or superscripts of a, b, c and d is 5.
Which means we need to check for those coefficients only which make the sum of powers of a, b, c, d as 5.
So we will proceed further by using the binomial expansion formula for ${(p + q)^n}$
${}^n{C_r}{p^r}{q^{n - r}}$where ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
So, for ${(x + y)^{10}}$=
${}^{10}{C_5}{x^5}{y^5}$
Substituting the values of x and y, we get
$ \Rightarrow {}^{10}{C_5}{(a - b)^5}{(d - c)^5}$
Now applying binomial expansion for both the terms again as per the required exponents of a, b , c, d ,we get
$ \Rightarrow {}^{10}{C_5} \cdot \left[ {{}^5{C_2} \cdot {a^2} \cdot {{( - b)}^3}} \right] \cdot \left[ {{}^5{C_1}{{( - c)}^4} \cdot d} \right]$
Rearranging the terms we get
$\Rightarrow - {}^{10}{C_5} \cdot {}^5{C_2} \cdot {}^5{C_1} \cdot {a^2} \cdot {b^3} \cdot {c^4} \cdot d$
Now , the coefficients of ${a^2}{b^3}{c^4}d$ will be,
$
= - {}^{10}{C_5}.{}^5{C_2}.{}^5{C_1} \\
= - 12600 \\
$
So , the coefficient of ${a^2}{b^3}{c^4}d$ is\[ - 12600\].
Note: Such similar questions can be solved by the same approach used as above. Attention must be given while expanding the binomial expression as any wrong input might lead to vague or incorrect answers. Also ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ value can never be negative, but the coefficients can be due to the nature of variable.
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