Permutations with Repeated Elements-Problems 15 and 16: The...

Permutations with Repeated Elements-Problems 15 and 16: The word CARRIER has seven letters. But there are fewer than $7 !$ permutations, because in any arrangement of these seven letters the three $R$ 's are interchangeable. If these $R$ 's were distinguishable, there would be $3 !$, or $6,$ ways of arranging them. This implies that only $\frac{1}{6}$ (that is, $\frac{1}{3 !}$ ) of the $7 !$ permutations are actually different. So the number of permutations is $$ \frac{7 !}{3 !}=840 $$ There are four $I^{\prime}$ 's, four $S$ 's, and two $P$ 's in the word MISSISSIPPI, so the number of different permutations of its letters is $$ \frac{111}{4 ! 4 ! 2 !}=34,650 $$ Find the number of different permutations of the letters in each word. a. FREELY b. BUBBLES c. LILLY d. MISSISSAUGA e. HONOLULU f. HAWAIIAN

Permutations with Repeated Elements-Problems 15 and 16: The word CARRIER has seven letters. But there are fewer than $7 !$ permutations, because in any arrangement of these seven letters the three $R$ 's are interchangeable. If these $R$ 's were distinguishable, there would be $3 !$, or $6,$ ways of arranging them. This implies that only $\frac{1}{6}$ (that is, $\frac{1}{3 !}$ ) of the $7 !$ permutations are actually different. So the number of permutations is
$$
\frac{7 !}{3 !}=840
$$
There are four $I^{\prime}$ 's, four $S$ 's, and two $P$ 's in the word MISSISSIPPI, so the number of different permutations of its letters is
$$
\frac{111}{4 ! 4 ! 2 !}=34,650
$$
Find the number of different permutations of the letters in each word.
a. FREELY
b. BUBBLES
c. LILLY
d. MISSISSAUGA
e. HONOLULU
f. HAWAIIAN

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Key Concepts

Permutation

A permutation is an arrangement of objects in a specific order. In combinatorics, counting the number of possible arrangements often involves calculating factorials, which denote the product of all positive integers up to a specified number.

Factorial

The factorial function (denoted by n!) is the product of all positive integers from 1 to n. It is fundamental in counting problems, especially in determining the total number of ways to arrange a set of distinct objects.

Permutations with Repeated Elements

When some items in a set are identical, the straightforward permutation calculation (n!) overcounts arrangements because swapping identical items does not produce a new, distinguishable arrangement. To correct for this, the total is divided by the factorials of the counts of each set of repeated items.

Overcounting Correction

The process of dividing by the factorial of the number of repetitions adjusts for overcounting. This technique ensures that arrangements that are identical due to the interchangeability of repeated items are not counted multiple times.

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