Had the first term been a variable I would have proceeded by finding possible solutions of equation $a+b+c+d+e=5$, but the one in first variable may mean that certain solutions would be identical, I could not solve this problem.
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http://en.wikipedia.org/wiki/Multinomial_theorem#Number_of_multinomial_coefficients – lab bhattacharjee Jan 12 '14 at 10:17
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Binomial theorem... http://en.wikipedia.org/wiki/Binomial_theorem to be used for more variables. Previous comments is maybe more useful... – Umberto Jan 12 '14 at 10:17
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And doing the simple work of expanding it, the result is $x^5+5 x^4 y+5 x^4 z+5 x^4+10 x^3 y^2+20 x^3 y z+20 x^3 y+10 x^3 z^2+20 x^3 z+10 x^3+10 x^2 y^3+30 x^2 y^2 z+30 x^2 y^2+30 x^2 y z^2+60 x^2 y z+30 x^2 y+10 x^2 z^3+30 x^2 z^2+30 x^2 z+10 x^2+5 x y^4+20 x y^3 z+20 x y^3+30 x y^2 z^2+60 x y^2 z+30 x y^2+20 x y z^3+60 x y z^2+60 x y z+20 x y+5 x z^4+20 x z^3+30 x z^2+20 x z+5 x+y^5+5 y^4 z+5 y^4+10 y^3 z^2+20 y^3 z+10 y^3+10 y^2 z^3+30 y^2 z^2+30 y^2 z+10 y^2+5 y z^4+20 y z^3+30 y z^2+20 y z+5 y+z^5+5 z^4+10 z^3+10 z^2+5 z+1$, so the result you are searcing is 56 (if I am not wrong) – Umberto Jan 12 '14 at 10:19
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Ok, now you corrected your answer. Before the power was 5. Now is 54. My previous comments does not apply anymore. – Umberto Jan 12 '14 at 10:25
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Evaluating $(w+x+y+z)^5$ at $w=1$ doesn't change the number of terms, since the $a$ in $w^ax^by^cz^d$ is determined by $b,c,d$ (from $a+b+c+d=5$). | Hey Yiorgos S. Smyrlis and freak_warrior, you can't just change the numbers in a problem because you say so. Rolling back. – anon Jan 12 '14 at 10:25
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@freak_warrior and YiorgosS.Smyrlis: Editing is good, modifying the math content is bad. Please be more careful. – Did Jan 12 '14 at 10:27
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By the way. Some research done :-) When the power is $54$ and we have $4$ terms (as stated below) the result is 1463 terms ;-) Someone care to list them all? Just kidding... – Umberto Jan 12 '14 at 16:18
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If the first term had been a variable you would have been looking at $(w+x+y+z)^5$, i.e. at terms of the form $w^ax^by^cz^d$ where $a+b+c+d=5$ with $a,b,c,d$ non-negative integers (note, no $e$).
Instead you are looking at terms of the form $x^by^cz^d$ where $b+c+d \le 5$ with $b,c,d$ non-negative integers. But this has exactly the same number of solutions ${5+4-1 \choose 5}=56$: any solution to the former clearly gives a solution to the latter, and any solution to the latter gives a solution to the former with $a=5-b-c-d$.
Henry
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1It is still $5$ objects into $4$ bins, one bin happens to be where the exponent of $1$ goes. Using the "stars and bars" method, $$ \underbrace{\star}{\text{exponent of $1$}}|\underbrace{\star\star}{\text{exponent of $x$}}|\underbrace{\star}{\text{exponent of $y$}}|\underbrace{\star}{\text{exponent of $z$}} $$ is $\binom{5+4-1}{5}$. – robjohn Jan 12 '14 at 10:49