How many permutations of six 0's, five 1's and four 2's have the first 0 preceding the first 1?

Question

Given are six 0's,five 1's and four 2's. Consider all possible permutations of all these numbers. How many permutations have the first 0 preceding the first 1.

So I solved it like this

I took 011111 and chose six positions for them that's $15C6$ and from all the chosen positions first position will be occupied by 0 and rest by all the 1's. Then I can arrange the remaining numbers in $$\frac{9!}{5!4!}$$ ways. Multiplying both should give me the answer but my book mentions $15C4 \cdot 10C5$.

I don't know what I am missing . Any help will be appreciated

Answer 1

There may be many different approaches. Here is one:

• If the question was Given six $0$s and five $1$s, consider all possible permutations of all these numbers. How many permutations have the first $0$ preceding the first $1$? then you might say there are ${11 \choose 6}$ possibilities of which ${10 \choose 5}$ have the first number being $0$

• but in your question we also need to take account of the four $2$s, and their positions can be chosen in ${15 \choose 4}$ different ways

• so overall there are ${11 \choose 6}{15 \choose 4}$ distinct permutations, of which ${10 \choose 5}{15 \choose 4}$ have the first $0$ preceding the first $1$, as the book seems to say

Your approach will double-count many patterns, such as $000000222211111$, where it is not clear which of the $0$s is the special one from your $011111$

Answer 2

What you are missing is double counting. If the permutation starts $0011111$ you count it twice, once with the first $0$ as part of your $011111$ and once with the second being part of your $011111$.