To solve this problem, we need to determine how many permutations of the word "COMBINATORICS" exist where all the occurrences of 'O' come before all the occurrences of 'I'.

Step 1: Analyze the structure of the word

The word "COMBINATORICS" has 12 letters in total, and the distribution of letters is as follows:

• C: 2
• O: 2
• M: 1
• B: 1
• I: 2
• N: 1
• A: 1
• T: 1
• R: 1
• S: 1

Step 2: Factorial formula for permutations with identical letters

The total number of distinct permutations of the letters in the word "COMBINATORICS" (without any restrictions) is given by:

$\text{Total permutations} = \frac{12!}{2! \times 2! \times 2!}$

where $12!$ accounts for all 12 letters, and the $2!$s account for the repeated C's, O's, and I's. This gives the total number of permutations without any conditions.

Step 3: Apply the restriction (O comes before I)

Now, we are only interested in the cases where both 'O's appear before both 'I's.

Since there are 4 positions available for the 2 O's and 2 I's, the number of ways to place the O's and I's without any restriction is:

$\binom{4}{2} = \frac{4!}{2!(4-2)!} = 6$

Out of these 6 arrangements, only 1 of them satisfies the condition that both O's come before both I's (i.e., the arrangement "OOII").

Thus, the fraction of the total permutations where 'O' comes before 'I' is:

$\frac{1}{6}$

Step 4: Calculate the final result

To find the number of permutations where all 'O's are before all 'I's, we multiply the total number of permutations by $\frac{1}{6}$:

$\text{Permutations with O before I} = \frac{1}{6} \times \frac{12!}{2! \times 2! \times 2!}$

Now we calculate the factorials:

$12! = 479,001,600$ $2! = 2$ $\frac{12!}{2! \times 2! \times 2!} = \frac{479,001,600}{2 \times 2 \times 2} = \frac{479,001,600}{8} = 59,875,200$

Finally, multiplying by $\frac{1}{6}$:

$\text{Permutations with O before I} = \frac{59,875,200}{6} = 9,979,200$

Final Answer:

The number of permutations of the word "COMBINATORICS" where all 'O's are placed before all 'I's is 9,979,200.

Would you like a detailed breakdown of any particular step? Here are a few related questions you might explore:

1. How many total distinct permutations are there for "COMBINATORICS" without any restrictions?
2. What if we wanted all 'I's to come before all 'O's instead?
3. How many permutations of "COMBINATORICS" are there if no restrictions are applied?
4. What if the restriction was that all 'C's come before all 'M's instead?
5. How does the number of repeated letters affect the total number of permutations?

Tip: When dealing with permutations involving restrictions, consider how the restricted elements can be ordered and then apply the corresponding fraction to the total number of permutations.