The coefficient of a^{10} b^7 c^3 in the expa | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The general term in the expansion of $(bc + ca + ab)^{10}$ is given by the multinomial theorem as: $\frac{10!}{n_1! n_2! n_3!} (bc)^{n_1} (ca)^{n_

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(C) The general term in the expansion of $(bc + ca + ab)^{10}$ is given by the multinomial theorem as: $\frac{10!}{n_1! n_2! n_3!} (bc)^{n_1} (ca)^{n_2} (ab)^{n_3} = \frac{10!}{n_1! n_2! n_3!} a^{n_2+n_3} b^{n_1+n_3} c^{n_1+n_2}$, where $n_1 + n_2 + n_3 = 10$.We need the coefficient of $a^{10} b^7 c^3$. Comparing the powers, we have:$n_2 + n_3 = 10$$n_1 + n_3 = 7$$n_1 + n_2 = 3$Adding these equations: $2(n_1 + n_2 + n_3) = 20$, which is consistent with $n_1 + n_2 + n_3 = 10$.Solving for $n_1, n_2, n_3$:From $n_1 + n_2 + n_3 = 10$ and $n_2 + n_3 = 10$, we get $n_1 = 0$.Substituting $n_1 = 0$ into $n_1 + n_2 = 3$, we get $n_2 = 3$.Substituting $n_2 = 3$ into $n_2 + n_3 = 10$, we get $n_3 = 7$.The coefficient is $\frac{10!}{0! 3! 7!} = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} = 10 	imes 3 	imes 4 = 120$.

The coefficient of $a^{10} b^7 c^3$ in the expansion of $(bc + ca + ab)^{10}$ is

Similar Questions

The last digit of the number $7^{886}$ is

The number $111...1$ ($91$ times) is

The sum of the divisors of $2^5 \times 3^4 \times 5^2$ is

The coefficient of $x^{10}$ in the expansion of $(x+\frac{2}{x}-5)^{12}$ is

It is given that the number $43361$ can be written as a product of $two$ distinct prime numbers $p_1$ and $p_2$. Further,assume that there are $42900$ numbers which are less than $43361$ and are coprime to it. Then,the value of $p_1+p_2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module