How to Count Arrangements of the Letters in SEVENTEEN

Expert intro: This is a classic repeated-letter counting problem. The main trick is to treat the repeated E's and N's carefully, then split the work into a position-based count and a gap-based count.

Detailed walkthrough

We use the letters in SEVENTEEN:

• 4 E's
• 2 N's
• 1 S, 1 V, 1 T

So there are 9 letters in total.

(i) One of the E's is in the centre with 4 letters on either side

Fix an E in the middle. That leaves 8 letters to arrange:

• 3 E's
• 2 N's
• S, V, T

Number of arrangements:

$\frac{8!}{3!\,2!} = \frac{40320}{12} = 3360$

(ii) No E is next to another E

First arrange the 5 non-E letters:

• S, V, T, N, N

Number of ways:

$\frac{5!}{2!} = 60$

These 5 letters create 6 gaps:

$\_\,L\,\_\,L\,\_\,L\,\_\,L\,\_\,L\,\_$

To keep all E's separated, choose 4 of these 6 gaps for the E's:

$\binom{6}{4} = 15$

So the total number of arrangements is

$60 \times 15 = 900$

Answers

• (i) 3360
• (ii) 900

💡 Pitfall guide

A common mistake is to divide by $4!$ for the E's in part (ii). That would only be valid if the E's were arranged freely without any spacing condition. Here the key step is first placing the non-E letters, then using the gaps between them.

🔄 Real-world variant

If the condition changed to no two E's adjacent, but one letter such as N were also required to stay apart from another specific letter, the gap method would still work. You would just count the base arrangement first, then choose slots that satisfy all spacing restrictions. If the centre letter were not fixed as E in part (i), the count would change because the middle position could be occupied by a different letter.

🔍 Related terms

repeated letters, gap method, multiset permutation