Answer
Verified
425.7k+ views
Hint: In order to solve such a question of expansion and finding its coefficient first assume the coefficient as some unknown variable. Use permutation to find the number of such arrangements of a, b& c in order to find the unknown variable.
Given expansion is ${\left( {1 + a - b + c} \right)^9}$
Let the coefficient of ${a^3}{b^4}c$ is $\lambda $ .
Since the expansion has an overall power of 9. So we know that the sum of overall power of each term of expansion must be 9.
In order to make the overall power of the given term as 9 let us modify the term as follows
${a^3}{b^4}c = {1^1}{a^3}{b^4}{c^1}$
Now we have 9 variables some of which are repeating in the given term
$\left\{ {3\left( a \right),4\left( b \right),1\left( c \right),1\left( 1 \right)} \right\}$
Since the coefficient of each term within the expansion term is 1.
So the coefficient of the given term will be the number of ways the 9 variables can be arranged.
Now the number of different arrangements of $\left\{ {3\left( a \right),4\left( b \right),1\left( c \right),1\left( 1 \right)} \right\}$ are:
$
\lambda = \dfrac{{\left( {{\text{total number of variables}}} \right)!}}{{\left( {{\text{number of }}a} \right)! \times \left( {{\text{number of }}b} \right)! \times \left( {{\text{number of }}c} \right)!}} \\
\lambda = \dfrac{{9!}}{{3! \times 4! \times 1!}} \\
\lambda = \dfrac{{9!}}{{3! \times 4!}}\left[ {\because 1! = 1} \right] \\
$
Hence, the coefficient of ${a^3}{b^4}c$ in the expansion of ${\left( {1 + a - b + c} \right)^9}$ is $\dfrac{{9!}}{{3! \times 4!}}$ .
And option D is the correct option.
Note: In order to solve such questions containing a larger power in the expansion part it is easier to solve the problem with the help of permutation and combination, but the coefficient of terms inside the expansion term must be taken with greater care. This question can also be solved by the help of binomial expansion and finding the coefficient of the term particularly.
Given expansion is ${\left( {1 + a - b + c} \right)^9}$
Let the coefficient of ${a^3}{b^4}c$ is $\lambda $ .
Since the expansion has an overall power of 9. So we know that the sum of overall power of each term of expansion must be 9.
In order to make the overall power of the given term as 9 let us modify the term as follows
${a^3}{b^4}c = {1^1}{a^3}{b^4}{c^1}$
Now we have 9 variables some of which are repeating in the given term
$\left\{ {3\left( a \right),4\left( b \right),1\left( c \right),1\left( 1 \right)} \right\}$
Since the coefficient of each term within the expansion term is 1.
So the coefficient of the given term will be the number of ways the 9 variables can be arranged.
Now the number of different arrangements of $\left\{ {3\left( a \right),4\left( b \right),1\left( c \right),1\left( 1 \right)} \right\}$ are:
$
\lambda = \dfrac{{\left( {{\text{total number of variables}}} \right)!}}{{\left( {{\text{number of }}a} \right)! \times \left( {{\text{number of }}b} \right)! \times \left( {{\text{number of }}c} \right)!}} \\
\lambda = \dfrac{{9!}}{{3! \times 4! \times 1!}} \\
\lambda = \dfrac{{9!}}{{3! \times 4!}}\left[ {\because 1! = 1} \right] \\
$
Hence, the coefficient of ${a^3}{b^4}c$ in the expansion of ${\left( {1 + a - b + c} \right)^9}$ is $\dfrac{{9!}}{{3! \times 4!}}$ .
And option D is the correct option.
Note: In order to solve such questions containing a larger power in the expansion part it is easier to solve the problem with the help of permutation and combination, but the coefficient of terms inside the expansion term must be taken with greater care. This question can also be solved by the help of binomial expansion and finding the coefficient of the term particularly.
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Write an application to the principal requesting five class 10 english CBSE
Difference Between Plant Cell and Animal Cell
a Tabulate the differences in the characteristics of class 12 chemistry CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
Discuss what these phrases mean to you A a yellow wood class 9 english CBSE
List some examples of Rabi and Kharif crops class 8 biology CBSE