Sum of coefficients in the expansion of $(x + 2 y + z)^{10}$ is

2^{10}

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3^{10}

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4^{9}

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4^{10}

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Solution

The correct option is D $4^{10}$
Substituting $x = 1$ , $y = 1$ , $z = 1$ in the above expression, we get the sum of the binomial coefficients as
$(x + 2 y + z)^{10}$
$= (1 + 2 (1) + 1)^{10}$ (substituting $x = 1, y = 1, z = 1$ )
$= (4)^{10}$
Hence answer is D

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