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  4. The number of dissimilar terms in the expansion of (x + y)n is n + 1. Therefore number of dissimilar terms in the expansion of (x + y + z)12 is

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Q. The number of dissimilar terms in the expansion of $ (x + y)^n$ is $n + 1$. Therefore number of dissimilar terms in the expansion of $ (x + y + z)^{12}$ is

Binomial Theorem

Solution:

$(x + y + z)^{12} = x^{12} + ^{12}c_1x^{11}(y + z)^1$
$+ ^{12}c_2x^{10}\, (y + z)^2 + ..... + \,{}^{12}c_{12}(y + z)^{12}$
$\therefore $ total no. of terms $= 1+2 + 3 + 4 +..... + 13$
$= \frac{13}{2}\left(1+13\right) = 13 \times 7 = 91$