In expansion of (1 + x + y + z)^4 ratio of coefficient of x^2y, xy^2z, xyz are

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7.Binomial Theorem

normal

In the expansion of $(1 + x + y + z)^4$ the ratio of coefficient of $x^2y, xy^2z, xyz$ are

$1 : 1 : 2$

$2 : 1 : 1$

$1 : 2 : 1$

Not defined

Solution

${(1 + x + y + z)^4} \to $ coefficient of $x^{2} y, x y^{2} z, x y z$

$\left( {^4{c_1}{\,^3}{c_2}{\,^1}{c_1}} \right),\left( {{\,^4}{c_1}{\,^3}{c_2}{\,^1}{c_1}} \right),\left( {^4{c_1}{\,^3}{c_1}{\,^2}{c_1}{\,^1}{c_1}} \right)$

$12: 12: 24$

$1: 1: 2$

Standard 11

Mathematics

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hard

(JEE MAIN-2024)

In the expansion of $(1 + x + y + z)^4$ the ratio of coefficient of $x^2y, xy^2z, xyz$ are

Solution

Similar Questions

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