Counting STATISTICS arrangements with separated I letters

Expert intro: This is a permutations problem with repeated letters and a spacing condition. The key is to count the placements of the two I's first, then arrange the remaining letters while accounting for duplicates.

Detailed walkthrough

Count the letters first

The word STATISTICS has 10 letters:

• S appears 3 times
• T appears 3 times
• I appears 2 times
• A appears 1 time
• C appears 1 time

We need arrangements in which the two I’s are separated by exactly 3 letters. That means if one I is in position $k$, the other must be in position $k+4$.

Choose the positions for the I's

There are 10 positions in a row. The pair of positions must differ by 4.

Possible pairs are:

• $(1,5)$
• $(2,6)$
• $(3,7)$
• $(4,8)$
• $(5,9)$
• $(6,10)$

So there are 6 valid ways to place the two I’s.

Arrange the remaining letters

After placing the I’s, the remaining 8 positions are filled by the letters S, S, S, T, T, T, A, C.

The number of distinct permutations of these 8 letters is

$\dfrac{8!}{3!3!}$.

Compute it:

$\dfrac{40320}{6\cdot 6}=1120$.

Multiply the independent choices

Each valid placement of the I’s can be combined with any arrangement of the remaining letters, so the total number is

$6 \times 1120 = 6720$.

Final answer

$\boxed{6720}$

The main idea is to separate the spacing condition from the repeated-letter counting. Once the I’s are fixed, the rest is a standard multinomial permutation count.

💡 Pitfall guide

A frequent mistake is to interpret “separated by exactly 3 letters” as meaning the I’s are 3 positions apart, rather than 4 positions apart with three letters in between. That off-by-one error changes the answer completely. Another trap is forgetting that STATISTICS contains repeated letters, so using $8!$ alone would overcount arrangements that only differ by swapping identical S’s or T’s. When a word has repeated letters, the division by factorials is not optional; it is the correction that makes the count distinct arrangements only.

🔄 Real-world variant

If the question changed to “How many arrangements have the two I’s separated by exactly 2 letters?” then the valid I positions would be $(1,4),(2,5),(3,6),(4,7),(5,8),(6,9),(7,10)$, giving 7 placements instead of 6. The remaining eight letters would still be arranged in $\dfrac{8!}{3!3!}$ ways, so the total would become $7\cdot \dfrac{8!}{3!3!}=7840$. A different spacing rule changes only the first counting step, not the duplicate-letter arrangement formula.

🔍 Related terms

multinomial permutation, repeated letters, spacing condition