In the multinomial expansion of $(a+b+c)^5$ , why does "a" have $5 \choose 1$ positions?

Question

Also, why does b has $4\choose 2$ positions and c has $3 \choose 2$ positions? , i.e. $(a+b+c)^5=a^5+...+abbcc+babcc+cabcb+....$

Answer 1

your question is not clear and you say there are ${{5}\choose{1}}$ positions for $a$ which is not totally true, if you want to see the coefficient of $a^1$ then set $k_2=1$ in the following formula.

$$\left(a+b+c\right)^5=\left(\left(a+b\right)+c\right)^{5}$$$$=\sum_{k_{1}=0}^{5}{{5}\choose{k_1}}\color{red}{\left(a+b\right)^{k_{1}}}\left(c\right)^{\left(5-k_{1}\right)}=\sum_{k_{1}=0}^{5}\color{red}{\sum_{k_{2}=0}^{k_{1}}}{{5}\choose{k_1}}\color{red}{{{k_1}\choose{k_2}}\left(a\right)^{k_{2}}\left(b\right)^{\left(k_{1}-k_{2}\right)}}\left(c\right)^{\left(5-k_{1}\right)}$$

Then for exmaple if you you want to know the position of $a^5$ , it can be done by setting $k_2=5$ , the same strategy can be applied for $b^5,c^5$

Here I've just used binomial theorem and nothing more.

Answer 2

The multinomial theorem can come to the rescue: $$ (a+b+c)^n=\sum_{i+j+k=n}\binom{n}{i,j,k}a^ib^jc^k $$

where $$ \dbinom{n}{i,j,k} = \dfrac{n!}{i! \, j! \, k!} $$

Here $n=5$

$$ (a+b+c)^5 = \binom{5}{5, 0, 0}a^5b^0c^0 + \binom{5}{4, 1, 0}a^4b^1c^0 + \binom{5}{4, 0, 1}a^4b^0c^1 + \binom{5}{3, 2, 0}a^3b^2c^0 + \binom{5}{3, 1, 1}a^3b^1c^1 + \binom{5}{3, 0, 2}a^3b^0c^2 + ... + \binom{5}{1, 2, 2}a^1b^2c^2 + ... + \binom{5}{0, 0, 5}a^0b^0c^5 $$

$$ = \frac{5!}{5!0!0!}a^5 + \frac{5!}{4!1!0!}a^4b + \frac{5!}{4!0!1!}a^4c + \frac{5!}{3!2!0!}a^3b^2 + \frac{5!}{3!1!1!}a^3bc +\frac{5!}{3!0!2!}a^3c^2 + ... + \frac{5!}{1!2!2!}ab^2c^2 + ... + \frac{5!}{0!0!5!}c^5 $$

where 0! = 1.