Find the number of (i) combinations and (ii) permutations of the letters of the word ACCOUNTANCY taken 4 at a time.

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> [!TLDR] ACCOUNTANCY's 11 letters reduce to 7 distinct letters with repetitions A(2), C(3), N(2), O(1), U(1), T(1), Y(1); counting 4-letter combinations by repetition pattern gives 89 combinations and 1,422 permutations.

> [!FORMULA] With repeated letters, split combinations by pattern: all-different uses ${}^kC_4$ (k = number of distinct letters); "2 alike + 2 different" and similar patterns use combinations of the repeat-eligible letters. Each combination's permutations $=\dfrac{4!}{(	ext{repeat factorial})}$; total permutations $=\sum(	ext{combinations in a pattern})	imes(	ext{arrangements per pattern})$.

1. ACCOUNTANCY $=$ A,C,C,O,U,N,T,A,N,C,Y — 11 letters; counting repeats: A$	imes2$, C$	imes3$, N$	imes2$, and O, U, T, Y each once. So there are 7 distinct letters, and letters with $\ge2$ copies are $\{C,A,N\}$.

**Combinations of 4 letters (by pattern)**
2. All 4 different: choose 4 of the 7 distinct letters: ${}^7C_4=35$.
3. 2 alike + 2 different: choose the repeated letter from $\{C,A,N\}$: 3 ways; choose 2 different letters from the other 6 distinct letters: ${}^6C_2=15$; combinations $=3	imes15=45$.
4. 2 alike + 2 alike (two different pairs): choose 2 letters from $\{C,A,N\}$: ${}^3C_2=3$. …

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