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How many distinct arrangements can be made from:

-advisor

-woodworker

I think advisor would be 7! and woodworker would be 10! or does woodworker count differently because of the repeated letters?

Cackling Muse
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  • Do you mean the letters in those words? If so, I think woodworker would be less than 10! because of the repeats. How might you count the number of "over-counts" due to repeated letters? – Quin May 04 '21 at 04:02
  • Your question doesn't actually say so, but I'll assume each arrangement is meant to use all of the available letters. How many different $2$-letter arrangements can be made from the letters in the word "it"? How many different $2$-letter arrangements can me made from the letters in the word "aa"? Are those two numbers the same? Does that help you understand that the answer for "woodworker" has to be less than $10!$? – Robert Shore May 04 '21 at 04:09
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    Hmmm.....so woodworker has 10 letters, W=2, O=3, D=1, R=2, K=1, E=1...so would I be able to solve it with 10! / 3!2!2! – Cackling Muse May 04 '21 at 04:15

1 Answers1

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I'll do it just for 'woodworker', 'advisor' should become apparent after this.

Let's assume that each of the $\text{W}$s, $\text{R}$s and $\text{O}$s are different, and designate them as $\text{W}_1$ and $\text{W}_2$, $\text{R}_1$ and $\text{R}_2$ and as $\text{O}_1,\text{O}_2$ and $\text{O}_3$ respectively. Then the number of ways in which the letters

$$\text{W}_1\text{O}_1\text{O}_2\text{D}\text{W}_2\text{O}_3\text{R}_1\text{K}\text{E}\text{R}_2$$

can be permuted is just the number of ways in which $10$ distinct objects can be permuted, which is $10!$.

However, the letters which we repeated are in fact identical. Consider a simpler example: the letters $\text{AAB}$. Suppose that we've separated the $\text{A}$s into $\text{A}_1$ and $\text{A}_2$. The possible permutations become:

$$\text{A}_1\text{A}_2\text{B}$$ $$\text{A}_1\text{B}\text{A}_2$$ $$\text{A}_2\text{A}_1\text{B}$$ $$\text{A}_2\text{B}\text{A}_1$$ $$\text{B}\text{A}_1\text{A}_2$$ $$\text{B}\text{A}_2\text{A}_1$$

The key fact here is noticing that we've actually successfully found the various positions that the $\text{A}$s can take. Here, those positions are $\text{AAB}$, $\text{ABA}$ and $\text{BAA}$. In those positions, we now have to remove any sense of ordering that the $\text{A}$s have among themselves: since there are $2$, of them, the number of ways that the $\text{A}$s are ordered is $2!$. Indeed, we have overcounted $2!$ times. Then the answer becomes $\frac{3!}{2!}$.

Coming back to 'Woodworker,' there are two $\text{W}$s, two $\text{R}$s and three $\text{O}$s, which can be ordered amongst themselves in $2!$, $2!$ and $3!$ ways respectively. Removing the overcounts, our answer becomes

$$\frac{10!}{2!2!3!}$$

Lt. Commander. Data
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