How many words are there consisting of 3 D's and 7 U's? (Dividers and Units)

Question

I am studying combinatorics and was solving this problem, I ran into a doubt with this one and thus need help. The cause of my confusion is that I have two fold approaches in my head when I try to solve this and I am not able to feel confident which is right.
Approach 1
Assuming 3 D's to be indistinguishable among themselves and 7 U's to be indistinguishable among themselves. We have 10 letters in total with 3 of one kind, and 7 of the other. The answer becomes
$\frac{10!}{{3!}{7!}}$
Approach 2
I don't have much info to support my cause but a gut feeling tells me this problem might be the case of Bose-Einstein statistics where we have unordered sampling with replacement, and in that case the answer becomes
${7 + 3 - 1} \choose 3$

This problem is taken from George Martin's Book on Enumerative Combinatorics. Any help is really appreciated. Further, if you could share how do you decide if the assumption you make while solving combinatorics problem is correct or not. Thanks

Answer 1

Your Approach 2 of ${7+3-1 \choose 3}$ would be the correct answer to a slightly different question, such as how many ways to place the three indistinguishable $D$s into seven distinguishable locations

But here you in fact have eight distinguishable locations, namely:

• before the first $U$
• between the first and the second $U$s
• between the second and the third $U$s
• between the third and the fourth $U$s
• between the fourth and the fifth $U$s
• between the fifth and the sixth $U$s
• between the sixth and the seventh $U$s
• after the seventh $U$

making the actual number of ways of do this using Approach 2: ${8+3-1 \choose 3} = {10 \choose 3}$, the same number as the ${7+3 \choose 3} = {10 \choose 3}$ number found using Approach 1