Summation Translation

Question

I translated a sum of terms being multiplied into two summations being multiplied, but I'm not sure if the summation's I've made matches the sum I originally had.

Original Equation: $\frac{26!}{8!(26-8)!} + \frac{26!}{7!(26-7)!}(\frac{26!}{1!(26-1)!}) + \frac{26!}{6!(26-6)!}(\frac{26!}{2!(26-2)!}) + \frac{26!}{5!(26-5)!}(\frac{26!}{3!(26-3)!}) = 264,517,825$

My Summations: $\sum_{n=8}^5 \frac{26!}{n!(26-n)!} * \sum_{i=0}^3 \frac{26!}{i!(26-i)!}$

My calculator enters overflow mode when I try to check with that.

Answer 1

The original equation is $$\binom{26}{8}+\binom{26}{7}\cdot\binom{26}{1}+\binom{26}{6}\cdot\binom{26}{2}+\binom{26}{5}\cdot\binom{26}{3}$$

Expanding out your sum gives: $$\bigg[\binom{26}{1}+\binom{26}{2}+\binom{26}{3}\bigg]\bigg[\binom{26}{5}+\binom{26}{6}+\binom{26}{7}+\binom{26}{8}\bigg]$$ these quite clearly don't match, your summation is significantly larger.

To solve this, note that $\binom{26}{0}=1$, and so the original equation can be seen as: $$\binom{26}{8-0}\cdot\binom{26}{0}+\binom{26}{8-1}\cdot\binom{26}{1}+\binom{26}{8-2}\cdot\binom{26}{2}+\binom{26}{8-3}\cdot\binom{26}{3}$$

And we can write this in sum notation as: $$\sum_{n=0}^3 \binom{26}{8-n}\cdot\binom{26}{n}=\sum_{n=0}^3\frac{26!^2}{n!\cdot (8-n)!\cdot (26-n)!\cdot (18+n)!}$$

Answer 2

The two are not equal. You started with something of the form

$$a_1b_1+a_2b_2+a_3b_3+a_4b_4\tag{1}$$

and changed it to

$$(a_1+a_2+a_3+a_4)(b_1+b_2+b_3+b_4)\,.\tag{2}$$

If you multiply out $(2)$, you get all $4\cdot4=16$ terms of the form $a_ib_j$ with $1\le i,j\le 4$, not just the ones with $i=j$ that appear in $(1)$. You can see very easily that the two aren’t the same by letting all $8$ of the variables be $1$: then $(1)$ evaluates to $4$, but $(2)$ evaluates to $4\cdot 4=16$.

Answer 3

You need only one sum and not a product of two sums.

$$ \sum _{n=5} ^{8} ( \frac{26!}{n! (26-n)!} . \frac{26!}{(8-n)!(26-8+n)!}) $$