Number of ways to arrange the letters BANANAS
$begingroup$
I was in class, and the teacher told us, that the number of ways to re-arrange the letters in BANANAS is $frac{7!}{2!3!)}$. I understand where the $2, 3$ and $7$ are coming from (the number of $N$'s, the number of $A$'s, and the total number of letters respectively), and why we are using factorials (the number of ways to arrange $n$ things is $n!$), but I still feel like I am missing something crucial.
I suppose, why is it that we are dividing by $2!$ and $3!$?
Why is it that $2!$ and $3!$ are being multiplied together?
combinatorics permutations
edited Dec 5 '18 at 16:57
user3482749
4,206919
asked Dec 5 '18 at 16:20
nessness
375
$endgroup$
add a comment |
$begingroup$
I was in class, and the teacher told us, that the number of ways to re-arrange the letters in BANANAS is $frac{7!}{2!3!)}$. I understand where the $2, 3$ and $7$ are coming from (the number of $N$'s, the number of $A$'s, and the total number of letters respectively), and why we are using factorials (the number of ways to arrange $n$ things is $n!$), but I still feel like I am missing something crucial.
I suppose, why is it that we are dividing by $2!$ and $3!$?
Why is it that $2!$ and $3!$ are being multiplied together?
combinatorics permutations
edited Dec 5 '18 at 16:57
user3482749
4,206919
asked Dec 5 '18 at 16:20
nessness
375
$endgroup$
$begingroup$
We are dividing by 2! and 3! simply because all 3 A's are alike letters and 2 N's are alike. They can be shuffled in 3! and 2! ways respectively.
$endgroup$
– Love Invariants
Dec 5 '18 at 16:24
1
$begingroup$
Try with a smaller case like $ANANA$. Write down all the arrangements. Their number should be $5!/2!/3!=10$
$endgroup$
– Robert Z
Dec 5 '18 at 16:25
add a comment |
$begingroup$
I was in class, and the teacher told us, that the number of ways to re-arrange the letters in BANANAS is $frac{7!}{2!3!)}$. I understand where the $2, 3$ and $7$ are coming from (the number of $N$'s, the number of $A$'s, and the total number of letters respectively), and why we are using factorials (the number of ways to arrange $n$ things is $n!$), but I still feel like I am missing something crucial.
I suppose, why is it that we are dividing by $2!$ and $3!$?
Why is it that $2!$ and $3!$ are being multiplied together?
combinatorics permutations
edited Dec 5 '18 at 16:57
user3482749
4,206919
asked Dec 5 '18 at 16:20
nessness
375
$endgroup$
I was in class, and the teacher told us, that the number of ways to re-arrange the letters in BANANAS is $frac{7!}{2!3!)}$. I understand where the $2, 3$ and $7$ are coming from (the number of $N$'s, the number of $A$'s, and the total number of letters respectively), and why we are using factorials (the number of ways to arrange $n$ things is $n!$), but I still feel like I am missing something crucial.
I suppose, why is it that we are dividing by $2!$ and $3!$?
Why is it that $2!$ and $3!$ are being multiplied together?
combinatorics permutations
combinatorics permutations
edited Dec 5 '18 at 16:57
user3482749
4,206919
asked Dec 5 '18 at 16:20
nessness
375
edited Dec 5 '18 at 16:57
user3482749
4,206919
asked Dec 5 '18 at 16:20
nessness
375
edited Dec 5 '18 at 16:57
user3482749
4,206919
edited Dec 5 '18 at 16:57
user3482749
4,206919
edited Dec 5 '18 at 16:57
user3482749
4,206919
4,206919
asked Dec 5 '18 at 16:20
nessness
375
asked Dec 5 '18 at 16:20
nessness
375
asked Dec 5 '18 at 16:20
nessness
375
375
$begingroup$
We are dividing by 2! and 3! simply because all 3 A's are alike letters and 2 N's are alike. They can be shuffled in 3! and 2! ways respectively.
$endgroup$
– Love Invariants
Dec 5 '18 at 16:24
1
$begingroup$
Try with a smaller case like $ANANA$. Write down all the arrangements. Their number should be $5!/2!/3!=10$
$endgroup$
– Robert Z
Dec 5 '18 at 16:25
add a comment |
$begingroup$
We are dividing by 2! and 3! simply because all 3 A's are alike letters and 2 N's are alike. They can be shuffled in 3! and 2! ways respectively.
$endgroup$
– Love Invariants
Dec 5 '18 at 16:24
1
$begingroup$
Try with a smaller case like $ANANA$. Write down all the arrangements. Their number should be $5!/2!/3!=10$
$endgroup$
– Robert Z
Dec 5 '18 at 16:25
$begingroup$
We are dividing by 2! and 3! simply because all 3 A's are alike letters and 2 N's are alike. They can be shuffled in 3! and 2! ways respectively.
$endgroup$
– Love Invariants
Dec 5 '18 at 16:24
$begingroup$
We are dividing by 2! and 3! simply because all 3 A's are alike letters and 2 N's are alike. They can be shuffled in 3! and 2! ways respectively.
$endgroup$
– Love Invariants
Dec 5 '18 at 16:24
1
1
$begingroup$
Try with a smaller case like $ANANA$. Write down all the arrangements. Their number should be $5!/2!/3!=10$
$endgroup$
– Robert Z
Dec 5 '18 at 16:25
$begingroup$
Try with a smaller case like $ANANA$. Write down all the arrangements. Their number should be $5!/2!/3!=10$
$endgroup$
– Robert Z
Dec 5 '18 at 16:25
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Assume that all the letter are different. For example, imagine that each $A$ and $N$ is colored differently.
Then the number of arrangements is $7!$. Imagine that you write down all of them.
Now return all colored letters to black, and note that there are many 'words' that are counted multiple times. Namely, every word is counted $2!3!$ times, exactly the number of ways to rearrange the $A$'s and $N$'s in it.
edited Dec 5 '18 at 17:17
AryanSonwatikar
40514
answered Dec 5 '18 at 16:24
ajotatxeajotatxe
53.8k23890
$endgroup$
add a comment |
$begingroup$
Consider making the $A$'s nd the $N$'s different by, say, putting a label on them, like so:
$$
BA_1N_1A_2N_2A_3S
$$
Then there truly are $7!$ ways to rearrange this word.
However, once we remove the labels, some of these $7!$ arrangements collapse into the same arrangement. Let's remove the labels on the $N$'s first. In that case, for instance,
$$
BA_1N_1A_2N_2A_3S\
BA_1N_2A_2N_1A_3S
$$
will become the same. In fact, all the $7!$ arrangements group together two by two in pairs which become the same after removing the labels from the $N$'s. The number of distinct words after removing the $N$ labels is therefore $frac{7!}{2}$.
Now let's look at these $frac{7!}{2}$ words, and remove the labels from the $A$'s as well. Again, many different arrangements will collapse into the same arrangement. For instance,
$$
BA_1NA_2NA_3S\
BA_1NA_3NA_2S\
BA_2NA_1NA_3S\
BA_2NA_3NA_1S\
BA_3NA_1NA_2S\
BA_3NA_2NA_1S
$$
will all become the same word after removing the labels. In fact, all the $frac{7!}{2}$ different arrangements will go together in groups of $6$ where all the words in one group become equal after removing the label. There are $frac{7!}{2cdot 6}$ such groups, so that's the number of distinguishable words after removing all the labels from both the $N$'s and the $A$'s.
answered Dec 5 '18 at 16:30
ArthurArthur
113k7115197
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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$begingroup$
Assume that all the letter are different. For example, imagine that each $A$ and $N$ is colored differently.
Then the number of arrangements is $7!$. Imagine that you write down all of them.
Now return all colored letters to black, and note that there are many 'words' that are counted multiple times. Namely, every word is counted $2!3!$ times, exactly the number of ways to rearrange the $A$'s and $N$'s in it.
edited Dec 5 '18 at 17:17
AryanSonwatikar
40514
answered Dec 5 '18 at 16:24
ajotatxeajotatxe
53.8k23890
$endgroup$
add a comment |
$begingroup$
Assume that all the letter are different. For example, imagine that each $A$ and $N$ is colored differently.
Then the number of arrangements is $7!$. Imagine that you write down all of them.
Now return all colored letters to black, and note that there are many 'words' that are counted multiple times. Namely, every word is counted $2!3!$ times, exactly the number of ways to rearrange the $A$'s and $N$'s in it.
edited Dec 5 '18 at 17:17
AryanSonwatikar
40514
answered Dec 5 '18 at 16:24
ajotatxeajotatxe
53.8k23890
$endgroup$
add a comment |
$begingroup$
Assume that all the letter are different. For example, imagine that each $A$ and $N$ is colored differently.
Then the number of arrangements is $7!$. Imagine that you write down all of them.
Now return all colored letters to black, and note that there are many 'words' that are counted multiple times. Namely, every word is counted $2!3!$ times, exactly the number of ways to rearrange the $A$'s and $N$'s in it.
edited Dec 5 '18 at 17:17
AryanSonwatikar
40514
answered Dec 5 '18 at 16:24
ajotatxeajotatxe
53.8k23890
$endgroup$
Assume that all the letter are different. For example, imagine that each $A$ and $N$ is colored differently.
Then the number of arrangements is $7!$. Imagine that you write down all of them.
Now return all colored letters to black, and note that there are many 'words' that are counted multiple times. Namely, every word is counted $2!3!$ times, exactly the number of ways to rearrange the $A$'s and $N$'s in it.
edited Dec 5 '18 at 17:17
AryanSonwatikar
40514
answered Dec 5 '18 at 16:24
ajotatxeajotatxe
53.8k23890
edited Dec 5 '18 at 17:17
AryanSonwatikar
40514
edited Dec 5 '18 at 17:17
AryanSonwatikar
40514
edited Dec 5 '18 at 17:17
AryanSonwatikar
40514
40514
answered Dec 5 '18 at 16:24
ajotatxeajotatxe
53.8k23890
answered Dec 5 '18 at 16:24
ajotatxeajotatxe
53.8k23890
answered Dec 5 '18 at 16:24
ajotatxeajotatxe
53.8k23890
53.8k23890
add a comment |
add a comment |
$begingroup$
Consider making the $A$'s nd the $N$'s different by, say, putting a label on them, like so:
$$
BA_1N_1A_2N_2A_3S
$$
Then there truly are $7!$ ways to rearrange this word.
However, once we remove the labels, some of these $7!$ arrangements collapse into the same arrangement. Let's remove the labels on the $N$'s first. In that case, for instance,
$$
BA_1N_1A_2N_2A_3S\
BA_1N_2A_2N_1A_3S
$$
will become the same. In fact, all the $7!$ arrangements group together two by two in pairs which become the same after removing the labels from the $N$'s. The number of distinct words after removing the $N$ labels is therefore $frac{7!}{2}$.
Now let's look at these $frac{7!}{2}$ words, and remove the labels from the $A$'s as well. Again, many different arrangements will collapse into the same arrangement. For instance,
$$
BA_1NA_2NA_3S\
BA_1NA_3NA_2S\
BA_2NA_1NA_3S\
BA_2NA_3NA_1S\
BA_3NA_1NA_2S\
BA_3NA_2NA_1S
$$
will all become the same word after removing the labels. In fact, all the $frac{7!}{2}$ different arrangements will go together in groups of $6$ where all the words in one group become equal after removing the label. There are $frac{7!}{2cdot 6}$ such groups, so that's the number of distinguishable words after removing all the labels from both the $N$'s and the $A$'s.
answered Dec 5 '18 at 16:30
ArthurArthur
113k7115197
$endgroup$
add a comment |
$begingroup$
Consider making the $A$'s nd the $N$'s different by, say, putting a label on them, like so:
$$
BA_1N_1A_2N_2A_3S
$$
Then there truly are $7!$ ways to rearrange this word.
However, once we remove the labels, some of these $7!$ arrangements collapse into the same arrangement. Let's remove the labels on the $N$'s first. In that case, for instance,
$$
BA_1N_1A_2N_2A_3S\
BA_1N_2A_2N_1A_3S
$$
will become the same. In fact, all the $7!$ arrangements group together two by two in pairs which become the same after removing the labels from the $N$'s. The number of distinct words after removing the $N$ labels is therefore $frac{7!}{2}$.
Now let's look at these $frac{7!}{2}$ words, and remove the labels from the $A$'s as well. Again, many different arrangements will collapse into the same arrangement. For instance,
$$
BA_1NA_2NA_3S\
BA_1NA_3NA_2S\
BA_2NA_1NA_3S\
BA_2NA_3NA_1S\
BA_3NA_1NA_2S\
BA_3NA_2NA_1S
$$
will all become the same word after removing the labels. In fact, all the $frac{7!}{2}$ different arrangements will go together in groups of $6$ where all the words in one group become equal after removing the label. There are $frac{7!}{2cdot 6}$ such groups, so that's the number of distinguishable words after removing all the labels from both the $N$'s and the $A$'s.
answered Dec 5 '18 at 16:30
ArthurArthur
113k7115197
$endgroup$
add a comment |
$begingroup$
Consider making the $A$'s nd the $N$'s different by, say, putting a label on them, like so:
$$
BA_1N_1A_2N_2A_3S
$$
Then there truly are $7!$ ways to rearrange this word.
However, once we remove the labels, some of these $7!$ arrangements collapse into the same arrangement. Let's remove the labels on the $N$'s first. In that case, for instance,
$$
BA_1N_1A_2N_2A_3S\
BA_1N_2A_2N_1A_3S
$$
will become the same. In fact, all the $7!$ arrangements group together two by two in pairs which become the same after removing the labels from the $N$'s. The number of distinct words after removing the $N$ labels is therefore $frac{7!}{2}$.
Now let's look at these $frac{7!}{2}$ words, and remove the labels from the $A$'s as well. Again, many different arrangements will collapse into the same arrangement. For instance,
$$
BA_1NA_2NA_3S\
BA_1NA_3NA_2S\
BA_2NA_1NA_3S\
BA_2NA_3NA_1S\
BA_3NA_1NA_2S\
BA_3NA_2NA_1S
$$
will all become the same word after removing the labels. In fact, all the $frac{7!}{2}$ different arrangements will go together in groups of $6$ where all the words in one group become equal after removing the label. There are $frac{7!}{2cdot 6}$ such groups, so that's the number of distinguishable words after removing all the labels from both the $N$'s and the $A$'s.
answered Dec 5 '18 at 16:30
ArthurArthur
113k7115197
$endgroup$
Consider making the $A$'s nd the $N$'s different by, say, putting a label on them, like so:
$$
BA_1N_1A_2N_2A_3S
$$
Then there truly are $7!$ ways to rearrange this word.
However, once we remove the labels, some of these $7!$ arrangements collapse into the same arrangement. Let's remove the labels on the $N$'s first. In that case, for instance,
$$
BA_1N_1A_2N_2A_3S\
BA_1N_2A_2N_1A_3S
$$
will become the same. In fact, all the $7!$ arrangements group together two by two in pairs which become the same after removing the labels from the $N$'s. The number of distinct words after removing the $N$ labels is therefore $frac{7!}{2}$.
Now let's look at these $frac{7!}{2}$ words, and remove the labels from the $A$'s as well. Again, many different arrangements will collapse into the same arrangement. For instance,
$$
BA_1NA_2NA_3S\
BA_1NA_3NA_2S\
BA_2NA_1NA_3S\
BA_2NA_3NA_1S\
BA_3NA_1NA_2S\
BA_3NA_2NA_1S
$$
will all become the same word after removing the labels. In fact, all the $frac{7!}{2}$ different arrangements will go together in groups of $6$ where all the words in one group become equal after removing the label. There are $frac{7!}{2cdot 6}$ such groups, so that's the number of distinguishable words after removing all the labels from both the $N$'s and the $A$'s.
answered Dec 5 '18 at 16:30
ArthurArthur
113k7115197
answered Dec 5 '18 at 16:30
ArthurArthur
113k7115197
answered Dec 5 '18 at 16:30
ArthurArthur
113k7115197
answered Dec 5 '18 at 16:30
ArthurArthur
113k7115197
113k7115197
add a comment |
add a comment |
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$begingroup$
We are dividing by 2! and 3! simply because all 3 A's are alike letters and 2 N's are alike. They can be shuffled in 3! and 2! ways respectively.
$endgroup$
– Love Invariants
Dec 5 '18 at 16:24
1
$begingroup$
Try with a smaller case like $ANANA$. Write down all the arrangements. Their number should be $5!/2!/3!=10$
$endgroup$
– Robert Z
Dec 5 '18 at 16:25