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$

A.{\text{ }}\dfrac{{9!}}{{3!6!}} \\

B.{\text{ }}\dfrac{{9!}}{{4!5!}} \\

C.{\text{ }}\dfrac{{9!}}{{3!5!}} \\

D.{\text{ }}\dfrac{{9!}}{{3!4!}} \\

$

Answer
Verified

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.

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