How many arrangements are there for the $8$ letters of the word VISITING?

Question

How many arrangements are there for the $8$ letters of the word VISITING?

Solution:

So there are $\binom{8}{8}$, meaning 8 choose 8. But this gives a $8$ letter set not $8$ letter words, therefore

$\binom{8}{8}\times 8!$

1) Is this logic correct?
2) How do I evaluate $\binom{8}{8}\times 8!$

Answer 1

Hint: If all the letters were different, there would be $8!$ ways to arrange them.

But there are three $I$'s. You get a whole bunch of repeats in the $8!$ ways. How do you account for the three $I$'s? (How many times do you repeat each arrangement with a naïve counting?)

Spoiler:

$8!/3!$

Answer 2

If you have $8$ distinct objects then there are $8!$ factorial ways to arrange them. so for example there are $8!$ ways of arranging the symbols $VI_1SI_2TI_3NG$. However, in our case each $I_i$ simply has the value $I$, so for us $VI_1SI_2TI_3NG$ is exactly the same as $VI_2SI_1TI_3NG$, so we do not count them separately. Hence there are less than $8!$ distinct ways of arranging the letters of $VISITOR$. In fact for each arrangement of letters there are exactly $3!$ ways of permuting the $I_j's$ so that the arrangement is the same when we replace each $I_j$ with $I$. Hence we need to reduce the count of $8!$ by a factor of $3!$.