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Original Question:

Given $(x + y + z)^{50}$ . How many terms does it have ?

Solution:

The general term is $(x^{n1} + y^{n2} + z^{n3})^n$ coming from Multinomial Theorem.

Then $n\choose n1, n2, n3 $=$50\choose n1, n2, n3 $. Why this step is not considered the soltuion ?

  • Comment: I don't understand why these things happen. How should I know what step to do next from this point ?

n1 + n2 + n3 = r (coming from combination with repetitions)

n1 + n2 + n3 = 50

$\therefore$ n = 3 and r = 50

= $n+r-1\choose r$ = $3+50-1\choose 50$ = $52 \choose 50$ terms

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    I have problem understanding the logic of the solution provided by my textbook. Any clarification is appreciated. – 2017-02-19
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    The general term isn't $(x^{n_1}+y^{n_2}+z^{n_3})^n$ – 2017-02-19
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    It is the Multinomial theorem equation. https://en.wikipedia.org/wiki/Multinomial_theorem Giving us access to the powers n1, n2 n3 which can equal to the total power n, and then be used for n1 + n2 + n3 = r – 2017-02-19

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For the number of terms in $(x+y+z)^n$, think of it as:

Consider $x,y,z$ to be "distinct" boxes and $n$ be the number of "identical" balls. Now each term in the expansion, looks like $\text{(some coefficient)}x^py^qz^r$. And now think of these powers as the number of balls you can throw in each box. Therefore, total number of terms is equivalent to counting total number of ways you can throw in these balls in the boxes, which is easy to count, using the stars and bars method, which turns out to be $\binom{n+2}{2}$.

More generally, the number of terms in the expansion of $(x_1+x_2+\ldots+x_k)^n$ will be $\binom{n+k-1}{k-1}$

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    Because p, q, and r are all representing the balls that go inside distinct boxes of x. y. and z . The total number of balls is the sum of all the balls in the distinct boxes, which means the limiting factor, r, is the total number of balls, n. Am I right with my logic ? p + q + r = 50 (r, the limiting factor which is the total number of balls). – 2017-02-19
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    Yeah! You're right. Actually, I forgot to mention, in my answer what was $r$ in the final answer. Nevertheless, I've made the edits now. – 2017-02-19
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    Ok sounds good now. Thanks for clarification. – 2017-02-19
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    Pleasure was all mine! – 2017-02-19