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📅 May 2, 2026 👤 scraface

May 02, 2026

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📝 Recent Posts

Question: A home-schooled student is studying permutations of the word...
Solution: First, compute the total number of distinct arrangements of...
Total unrestricted arrangements:
\frac{6!}{3! \cdot 2! \cdot 1!} = \frac{720}{6 \cdot 2} = \frac{720}{1...
Now, subtract the number of arrangements where the two N's are adjacen...
Treat the two adjacent N's as a single block "NN". Then we are arrangi...
Number of such arrangements:
Therefore, the number of arrangements where the two N's are **not adja...
- 20 = 40
Thus, the number of valid distinct arrangements is:

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