The number of terms in the expansion of $$\left(x+y+z+w\right)^{10}$$
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https://en.wikipedia.org/wiki/Multinomial_theorem#Number_of_multinomial_coefficients – lab bhattacharjee Apr 18 '17 at 08:41
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Linking related questions: 1, 2, 3. – Frenzy Li Apr 18 '17 at 09:02
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Also related to this – Ng Chung Tak Apr 18 '17 at 10:42
2 Answers
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Each term is of the form $x^a y^b z^c w^d$ with $a+b+c+d=10$.
We have four elements, $x,y,z,w$, and we need to choose $10$ with repetition ignoring order.
This is called combination with repetition, or "multiset coefficient".
The required number is $\left(\dbinom{4}{10}\right) = \dbinom{4+10-1}{4-1} = \dbinom{13}3 = 286$.
DHMO
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We can start by considering smaller values for the exponent and seeing if we notice a pattern:
\begin{align}(x+y+z+w)^1&\text{ has } 4\text{ terms}\\ (x+y+z+w)^2&\text{ has } 10\text{ terms}\\ (x+y+z+w)^3&\text{ has } 20\text{ terms}\\ (x+y+z+w)^4&\text{ has } 35\text{ terms}\end{align}
These are triangle numbers, shunted by $1$, with the number of terms in $(x+y+z+w)^n$ calculated by $$\frac {(n+1)(n+2)(n+3)}{6}$$
Therefore, we have $(x+y+z+w)^{10}$ having $$\frac{11\times12\times13}6=286\text{ terms}$$
lioness99a
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solution is easy to understand but how you conclude the number of terms for 2, 3 and 4 power of expansion, as it is very tedious job or do you have any other approach? – user146551 Apr 18 '17 at 09:08
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@user146551 Personally, I used WolframAlpha for speed, however you can calculate them by hand - you square the expression, and then you just need to multiply the result by $(x+y+z+w)$ to get each subsequent exponent – lioness99a Apr 18 '17 at 09:11