Permutations

What makes it a permutation. A permutation of $n$ distinct objects taken $r$ at a time is an ordered selection and arrangement of $r$ of them. Two arrangements that use the same objects but place them in a different sequence count as different permutations. This is the single fact that separates permutations from combinations (which come later in the same chapter): order matters here.

The formula, from first principles. Think of filling $r$ positions in a row, one object per position, no repeats. The first position can be filled in $n$ ways (any of the $n$ objects). Once that choice is made, the second position has only $n-1$ objects left to choose from, the third has $n-2$, and so on, until the $r$-th position, which has $n-r+1$ choices left. By the multiplication principle these choices multiply together:

where: $n$ is the total number of distinct objects available, $r$ is how many of them are arranged in order, $n!$ (read "$n$ factorial") is $n(n-1)(n-2)\cdots 2\cdot 1$, and $^nP_r$ (also written $P(n,r)$) is the count of ordered arrangements. The special case $r = n$ — arranging all $n$ objects — gives $^nP_n = n!/0! = n!$, since $0! = 1$ by convention.

When repetition is allowed. If an object can be reused in a later position (say, digits in a code where repeats are permitted), every one of the $r$ positions independently has all $n$ objects available, so the count is simply $n^r$ — a different, usually much larger, number than $^nP_r$.

When the objects themselves repeat. Sometimes we are not choosing $r$ out of $n$ distinct objects but arranging all $n$ objects in a row, where several of them are indistinguishable copies of each other — like the three S's in STATISTICS. If $n$ objects split into groups of $p_1, p_2, \ldots, p_k$ identical objects (with $p_1 + p_2 + \cdots + p_k = n$), the number of visibly distinct arrangements is

where: each $p_i$ is the count of one repeated kind of object. The division is a correction: naively treating all $n$ objects as distinguishable and computing $n!$ overcounts, because swapping identical objects among themselves produces an arrangement that looks the same as before but was counted as new. Dividing by $p_1!$ removes the $p_1!$ ways to permute the first repeated group among itself, and likewise for each other group.

Problem. How many 3-letter arrangements (no letter repeated) can be made from the letters of the word FLOWER? Then, separately, in how many distinct ways can all the letters of the word STATISTICS be arranged?

Step 1. FLOWER has 6 distinct letters: F, L, O, W, E, R. Arranging 3 of them in order, with no repetition, is exactly $^nP_r$ with $n = 6$, $r = 3$: $^6P_3 = \dfrac{6!}{(6-3)!} = \dfrac{6!}{3!} = \dfrac{720}{6} = 120$.

Step 2. Cross-check FLOWER directly by the multiplication principle: first letter in 6 ways, second in 5 ways (one used), third in 4 ways, giving $6 \times 5 \times 4 = 120$. Both routes agree: 120 three-letter arrangements.

Step 3. Now STATISTICS. It has 10 letters total: S, T, A, T, I, S, T, I, C, S. Counting repeats: S appears 3 times, T appears 3 times, I appears 2 times, and A and C each appear once. Check the total: $3+3+2+1+1 = 10$.

Step 4. Since the letters are not all distinct, use $\dfrac{n!}{p_1!\,p_2!\cdots p_k!}$ with $n = 10$ and repeated-group counts $3, 3, 2$ (groups of size $1$ contribute $1! = 1$ and change nothing): $\dfrac{10!}{3!\,3!\,2!} = \dfrac{3628800}{6 \times 6 \times 2} = \dfrac{3628800}{72} = 50400$. The letters of STATISTICS can be arranged in 50,400 distinct ways, the division by $3!\,3!\,2!$ removing the arrangements that only differ by swapping identical S's, T's, or I's among themselves.

Competitive stretch

A tighter variant: how many of those STATISTICS arrangements keep all three S's together, treated as one block? Glue the three S's into a single unit; we now arrange $8$ items — that block plus the remaining 7 letters T, A, T, I, T, I, C — of which T repeats 3 times and I repeats 2 times. The count is $\dfrac{8!}{3!\,2!} = \dfrac{40320}{12} = 3360$. This "glue the repeated block" trick is a common extension once the base not-all-distinct formula is secure.

• Using $^nP_r$ when order does not actually matter. If the problem is about selecting a group (a committee, a hand of cards) and not sequencing it, permutations overcount — combinations are needed instead.
• Forgetting to divide out repeated objects. Treating STATISTICS-type words as if all letters were distinct and using plain $n!$ silently overcounts by a factor of $p_1!\,p_2!\cdots p_k!$.
• Mixing up "repetition allowed" with the standard case. With repetition, the count is $n^r$; without it, $n!/(n-r)!$ — these are different formulas for different situations, not interchangeable.
• Misreading $n$ and $r$ from the word problem, or slipping in the arithmetic of $(n-r)!$, especially when $n$ and $r$ must first be extracted from a real-world description rather than being handed directly.

Ask the tutor: Want to try a mixed set of permutation word problems — some with repetition, some with repeated letters — to build speed at spotting which formula applies?