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Find the number of distinct terms when expanded and collected in $(x+y+z)^{20}(x+y)^{15}$
How would I do this nicely?

I know that the first expansion has general algebraic term of
$$\frac{20}{b_1!b_2!b_3!} x^{b_1}y^{b_2}z^{b_3} $$ where $b_1+b_2+b_3 = 20, b_1 \geq 0$
and the second is $$\frac{15}{c_1 ! c_2 !} x^{c_1}y^{c_2} $$ where $c_1 + c_2 = 15, c_i \geq 0$.

Natash1
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2 Answers2

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$$=\sum_{r=0}^{20}\binom{20}r(x+y)^{15+r}z^{20-r}$$

Here each term is unique due to different exponents of $z$

Again, we know the number of terms of binomial expansion.

So, the no. Of distinct terms will be $$\sum_{r=0}^{20}(15+r+1)=?$$

lab bhattacharjee
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  • Thank you! I saw in some old solutions that this was equal to $\binom{37}2 - \binom{16}2$ without explanation except that they considered the complement: number of different terms there are. Would you be able to explain this? – Natash1 Nov 10 '17 at 01:06
  • Also, it would've been much tricker if the brackets were something like $(x+1/x + z)^{20} (x+z)^{15}$ right? Since some terms would simplify – Natash1 Nov 10 '17 at 01:07
  • @Natash1, That is same as $$\sum_{r=1}^{36}r-\sum_{r=1}^{15}r$$ – lab bhattacharjee Nov 10 '17 at 01:10
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Alternatively: denote $x+y=a$, then: $$(x+y+z)^{20}(x+y)^{15}=(a+z)^{20}a^{15}=a^{35}+20a^{34}z+\cdots+20a^{16}z^{19}+a^{15}z^{20}.$$ Note each term produces distinct terms, when $a=x+y$ is replaced back, thus: $$36+35+\cdots +16=\frac{16+36}{2}\cdot 21=546.$$

farruhota
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