In how many ways can the letters in the word “PROBABILITY” be arranged using the following restrcitions...
$begingroup$
In how many ways can the letters in the word "PROBABILITY" be arranged if the first letter must be "B" and the last letter cannot be an "O", "A", or "I"?
combinatorics permutations
edited Dec 11 '18 at 9:22
N. F. Taussig
44.1k93356
asked Dec 11 '18 at 7:44
Jacob PhelpsJacob Phelps
161
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closed as off-topic by Shailesh, Brahadeesh, egreg, Lord Shark the Unknown, Eric Wofsey Dec 12 '18 at 6:02
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shailesh, Brahadeesh, egreg, Lord Shark the Unknown, Eric Wofsey
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
In how many ways can the letters in the word "PROBABILITY" be arranged if the first letter must be "B" and the last letter cannot be an "O", "A", or "I"?
combinatorics permutations
edited Dec 11 '18 at 9:22
N. F. Taussig
44.1k93356
asked Dec 11 '18 at 7:44
Jacob PhelpsJacob Phelps
161
$endgroup$
closed as off-topic by Shailesh, Brahadeesh, egreg, Lord Shark the Unknown, Eric Wofsey Dec 12 '18 at 6:02
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shailesh, Brahadeesh, egreg, Lord Shark the Unknown, Eric Wofsey
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
In how many ways can the letters in the word "PROBABILITY" be arranged if the first letter must be "B" and the last letter cannot be an "O", "A", or "I"?
combinatorics permutations
edited Dec 11 '18 at 9:22
N. F. Taussig
44.1k93356
asked Dec 11 '18 at 7:44
Jacob PhelpsJacob Phelps
161
$endgroup$
In how many ways can the letters in the word "PROBABILITY" be arranged if the first letter must be "B" and the last letter cannot be an "O", "A", or "I"?
combinatorics permutations
combinatorics permutations
edited Dec 11 '18 at 9:22
N. F. Taussig
44.1k93356
asked Dec 11 '18 at 7:44
Jacob PhelpsJacob Phelps
161
edited Dec 11 '18 at 9:22
N. F. Taussig
44.1k93356
asked Dec 11 '18 at 7:44
Jacob PhelpsJacob Phelps
161
edited Dec 11 '18 at 9:22
N. F. Taussig
44.1k93356
edited Dec 11 '18 at 9:22
N. F. Taussig
44.1k93356
edited Dec 11 '18 at 9:22
N. F. Taussig
44.1k93356
44.1k93356
asked Dec 11 '18 at 7:44
Jacob PhelpsJacob Phelps
161
asked Dec 11 '18 at 7:44
Jacob PhelpsJacob Phelps
161
asked Dec 11 '18 at 7:44
Jacob PhelpsJacob Phelps
161
161
closed as off-topic by Shailesh, Brahadeesh, egreg, Lord Shark the Unknown, Eric Wofsey Dec 12 '18 at 6:02
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shailesh, Brahadeesh, egreg, Lord Shark the Unknown, Eric Wofsey
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Shailesh, Brahadeesh, egreg, Lord Shark the Unknown, Eric Wofsey Dec 12 '18 at 6:02
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shailesh, Brahadeesh, egreg, Lord Shark the Unknown, Eric Wofsey
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The word $PROBABILITY$ has eleven letters, of which $1$ is a $P$, $1$ is an $R$, $1$ is an O, $2$ are $B$s, $1$ is an $A$, $2$ are $I$s, $1$ is an $L$, $1$ is a $T$, and $1$ is an $I$. Hence, we have eleven positions to fill. Since the first letter must be a $B$, the first position can be filled in one way. Since there are nine distinct letters in the word $PROBABILITY$ and the last letter cannot be an $A$, $O$, or $I$, the last position can be filled in $6$ ways. Two of the remaining nine positions must be filled with $I$s, which can be done in $binom{9}{2}$ ways. Since we have already placed both $I$s and one of the $B$s, the remaining seven letters are distinct, so they can be placed in the remaining seven positions in $7!$ ways. Hence, the number of admissible arrangements of the letters of the word $PROBABILIY$ is
$$1 cdot 6 cdot binom{9}{2} cdot 7! = 6 cdot frac{9!}{2!7!} cdot 7! = 3 cdot 9! = 1,088,640$$
answered Dec 11 '18 at 9:33
N. F. TaussigN. F. Taussig
44.1k93356
$endgroup$
$begingroup$
can we use $^{n}C_{r}$ for choosing $r$ identical items
$endgroup$
– deleteprofile
Dec 11 '18 at 11:12
1
$begingroup$
We are choosing two positions for the $I$s. The positions are not identical.
$endgroup$
– N. F. Taussig
Dec 11 '18 at 12:31
add a comment |
$begingroup$
first letter should be $B$ (in $1$ way )and last letter can't be $O,A,I $ so, last letter can be $P,R,B,Y,L,T$ in $6$ ways and remaining $9$ letters can be arranged in between in $dfrac{9!}{2!}$ ways (division by $2!$ because of two identical I's)
so, total number of ways $=1times 6timesdfrac{9!}{2!}=1088640 $ ways
answered Dec 11 '18 at 10:53
deleteprofiledeleteprofile
1,155316
$endgroup$
add a comment |
$begingroup$
Solved 2 x 6 x (9!) = 4,354,560.
My error was using a 7 where the 6 was, forgot to consider the 2 at the beginning
answered Dec 11 '18 at 8:14
Jacob PhelpsJacob Phelps
161
$endgroup$
1
$begingroup$
Your answer is incorrect. You have not explained how you got your answer, so I am not sure what you did. Since the two $I$s are indistinguishable, you should have divided by $2$.
$endgroup$
– N. F. Taussig
Dec 11 '18 at 9:36
$begingroup$
Welcome to MathSE. The expectation at this site is that you include your attempt in the question itself, which you can do by clicking on the edit button at the bottom of your question. Explaining how you arrived at your answer helps users of this site identify any errors you may have made. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Dec 11 '18 at 9:40
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The word $PROBABILITY$ has eleven letters, of which $1$ is a $P$, $1$ is an $R$, $1$ is an O, $2$ are $B$s, $1$ is an $A$, $2$ are $I$s, $1$ is an $L$, $1$ is a $T$, and $1$ is an $I$. Hence, we have eleven positions to fill. Since the first letter must be a $B$, the first position can be filled in one way. Since there are nine distinct letters in the word $PROBABILITY$ and the last letter cannot be an $A$, $O$, or $I$, the last position can be filled in $6$ ways. Two of the remaining nine positions must be filled with $I$s, which can be done in $binom{9}{2}$ ways. Since we have already placed both $I$s and one of the $B$s, the remaining seven letters are distinct, so they can be placed in the remaining seven positions in $7!$ ways. Hence, the number of admissible arrangements of the letters of the word $PROBABILIY$ is
$$1 cdot 6 cdot binom{9}{2} cdot 7! = 6 cdot frac{9!}{2!7!} cdot 7! = 3 cdot 9! = 1,088,640$$
answered Dec 11 '18 at 9:33
N. F. TaussigN. F. Taussig
44.1k93356
$endgroup$
$begingroup$
can we use $^{n}C_{r}$ for choosing $r$ identical items
$endgroup$
– deleteprofile
Dec 11 '18 at 11:12
1
$begingroup$
We are choosing two positions for the $I$s. The positions are not identical.
$endgroup$
– N. F. Taussig
Dec 11 '18 at 12:31
add a comment |
$begingroup$
The word $PROBABILITY$ has eleven letters, of which $1$ is a $P$, $1$ is an $R$, $1$ is an O, $2$ are $B$s, $1$ is an $A$, $2$ are $I$s, $1$ is an $L$, $1$ is a $T$, and $1$ is an $I$. Hence, we have eleven positions to fill. Since the first letter must be a $B$, the first position can be filled in one way. Since there are nine distinct letters in the word $PROBABILITY$ and the last letter cannot be an $A$, $O$, or $I$, the last position can be filled in $6$ ways. Two of the remaining nine positions must be filled with $I$s, which can be done in $binom{9}{2}$ ways. Since we have already placed both $I$s and one of the $B$s, the remaining seven letters are distinct, so they can be placed in the remaining seven positions in $7!$ ways. Hence, the number of admissible arrangements of the letters of the word $PROBABILIY$ is
$$1 cdot 6 cdot binom{9}{2} cdot 7! = 6 cdot frac{9!}{2!7!} cdot 7! = 3 cdot 9! = 1,088,640$$
answered Dec 11 '18 at 9:33
N. F. TaussigN. F. Taussig
44.1k93356
$endgroup$
$begingroup$
can we use $^{n}C_{r}$ for choosing $r$ identical items
$endgroup$
– deleteprofile
Dec 11 '18 at 11:12
1
$begingroup$
We are choosing two positions for the $I$s. The positions are not identical.
$endgroup$
– N. F. Taussig
Dec 11 '18 at 12:31
add a comment |
$begingroup$
The word $PROBABILITY$ has eleven letters, of which $1$ is a $P$, $1$ is an $R$, $1$ is an O, $2$ are $B$s, $1$ is an $A$, $2$ are $I$s, $1$ is an $L$, $1$ is a $T$, and $1$ is an $I$. Hence, we have eleven positions to fill. Since the first letter must be a $B$, the first position can be filled in one way. Since there are nine distinct letters in the word $PROBABILITY$ and the last letter cannot be an $A$, $O$, or $I$, the last position can be filled in $6$ ways. Two of the remaining nine positions must be filled with $I$s, which can be done in $binom{9}{2}$ ways. Since we have already placed both $I$s and one of the $B$s, the remaining seven letters are distinct, so they can be placed in the remaining seven positions in $7!$ ways. Hence, the number of admissible arrangements of the letters of the word $PROBABILIY$ is
$$1 cdot 6 cdot binom{9}{2} cdot 7! = 6 cdot frac{9!}{2!7!} cdot 7! = 3 cdot 9! = 1,088,640$$
answered Dec 11 '18 at 9:33
N. F. TaussigN. F. Taussig
44.1k93356
$endgroup$
The word $PROBABILITY$ has eleven letters, of which $1$ is a $P$, $1$ is an $R$, $1$ is an O, $2$ are $B$s, $1$ is an $A$, $2$ are $I$s, $1$ is an $L$, $1$ is a $T$, and $1$ is an $I$. Hence, we have eleven positions to fill. Since the first letter must be a $B$, the first position can be filled in one way. Since there are nine distinct letters in the word $PROBABILITY$ and the last letter cannot be an $A$, $O$, or $I$, the last position can be filled in $6$ ways. Two of the remaining nine positions must be filled with $I$s, which can be done in $binom{9}{2}$ ways. Since we have already placed both $I$s and one of the $B$s, the remaining seven letters are distinct, so they can be placed in the remaining seven positions in $7!$ ways. Hence, the number of admissible arrangements of the letters of the word $PROBABILIY$ is
$$1 cdot 6 cdot binom{9}{2} cdot 7! = 6 cdot frac{9!}{2!7!} cdot 7! = 3 cdot 9! = 1,088,640$$
answered Dec 11 '18 at 9:33
N. F. TaussigN. F. Taussig
44.1k93356
answered Dec 11 '18 at 9:33
N. F. TaussigN. F. Taussig
44.1k93356
answered Dec 11 '18 at 9:33
N. F. TaussigN. F. Taussig
44.1k93356
answered Dec 11 '18 at 9:33
N. F. TaussigN. F. Taussig
44.1k93356
44.1k93356
$begingroup$
can we use $^{n}C_{r}$ for choosing $r$ identical items
$endgroup$
– deleteprofile
Dec 11 '18 at 11:12
1
$begingroup$
We are choosing two positions for the $I$s. The positions are not identical.
$endgroup$
– N. F. Taussig
Dec 11 '18 at 12:31
add a comment |
$begingroup$
can we use $^{n}C_{r}$ for choosing $r$ identical items
$endgroup$
– deleteprofile
Dec 11 '18 at 11:12
1
$begingroup$
We are choosing two positions for the $I$s. The positions are not identical.
$endgroup$
– N. F. Taussig
Dec 11 '18 at 12:31
$begingroup$
can we use $^{n}C_{r}$ for choosing $r$ identical items
$endgroup$
– deleteprofile
Dec 11 '18 at 11:12
$begingroup$
can we use $^{n}C_{r}$ for choosing $r$ identical items
$endgroup$
– deleteprofile
Dec 11 '18 at 11:12
1
1
$begingroup$
We are choosing two positions for the $I$s. The positions are not identical.
$endgroup$
– N. F. Taussig
Dec 11 '18 at 12:31
$begingroup$
We are choosing two positions for the $I$s. The positions are not identical.
$endgroup$
– N. F. Taussig
Dec 11 '18 at 12:31
add a comment |
$begingroup$
first letter should be $B$ (in $1$ way )and last letter can't be $O,A,I $ so, last letter can be $P,R,B,Y,L,T$ in $6$ ways and remaining $9$ letters can be arranged in between in $dfrac{9!}{2!}$ ways (division by $2!$ because of two identical I's)
so, total number of ways $=1times 6timesdfrac{9!}{2!}=1088640 $ ways
answered Dec 11 '18 at 10:53
deleteprofiledeleteprofile
1,155316
$endgroup$
add a comment |
$begingroup$
first letter should be $B$ (in $1$ way )and last letter can't be $O,A,I $ so, last letter can be $P,R,B,Y,L,T$ in $6$ ways and remaining $9$ letters can be arranged in between in $dfrac{9!}{2!}$ ways (division by $2!$ because of two identical I's)
so, total number of ways $=1times 6timesdfrac{9!}{2!}=1088640 $ ways
answered Dec 11 '18 at 10:53
deleteprofiledeleteprofile
1,155316
$endgroup$
add a comment |
$begingroup$
first letter should be $B$ (in $1$ way )and last letter can't be $O,A,I $ so, last letter can be $P,R,B,Y,L,T$ in $6$ ways and remaining $9$ letters can be arranged in between in $dfrac{9!}{2!}$ ways (division by $2!$ because of two identical I's)
so, total number of ways $=1times 6timesdfrac{9!}{2!}=1088640 $ ways
answered Dec 11 '18 at 10:53
deleteprofiledeleteprofile
1,155316
$endgroup$
first letter should be $B$ (in $1$ way )and last letter can't be $O,A,I $ so, last letter can be $P,R,B,Y,L,T$ in $6$ ways and remaining $9$ letters can be arranged in between in $dfrac{9!}{2!}$ ways (division by $2!$ because of two identical I's)
so, total number of ways $=1times 6timesdfrac{9!}{2!}=1088640 $ ways
answered Dec 11 '18 at 10:53
deleteprofiledeleteprofile
1,155316
edited Dec 11 '18 at 11:06
answered Dec 11 '18 at 10:53
deleteprofiledeleteprofile
1,155316
answered Dec 11 '18 at 10:53
deleteprofiledeleteprofile
1,155316
answered Dec 11 '18 at 10:53
deleteprofiledeleteprofile
1,155316
1,155316
add a comment |
add a comment |
$begingroup$
Solved 2 x 6 x (9!) = 4,354,560.
My error was using a 7 where the 6 was, forgot to consider the 2 at the beginning
answered Dec 11 '18 at 8:14
Jacob PhelpsJacob Phelps
161
$endgroup$
1
$begingroup$
Your answer is incorrect. You have not explained how you got your answer, so I am not sure what you did. Since the two $I$s are indistinguishable, you should have divided by $2$.
$endgroup$
– N. F. Taussig
Dec 11 '18 at 9:36
$begingroup$
Welcome to MathSE. The expectation at this site is that you include your attempt in the question itself, which you can do by clicking on the edit button at the bottom of your question. Explaining how you arrived at your answer helps users of this site identify any errors you may have made. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Dec 11 '18 at 9:40
add a comment |
$begingroup$
Solved 2 x 6 x (9!) = 4,354,560.
My error was using a 7 where the 6 was, forgot to consider the 2 at the beginning
answered Dec 11 '18 at 8:14
Jacob PhelpsJacob Phelps
161
$endgroup$
1
$begingroup$
Your answer is incorrect. You have not explained how you got your answer, so I am not sure what you did. Since the two $I$s are indistinguishable, you should have divided by $2$.
$endgroup$
– N. F. Taussig
Dec 11 '18 at 9:36
$begingroup$
Welcome to MathSE. The expectation at this site is that you include your attempt in the question itself, which you can do by clicking on the edit button at the bottom of your question. Explaining how you arrived at your answer helps users of this site identify any errors you may have made. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Dec 11 '18 at 9:40
add a comment |
$begingroup$
Solved 2 x 6 x (9!) = 4,354,560.
My error was using a 7 where the 6 was, forgot to consider the 2 at the beginning
answered Dec 11 '18 at 8:14
Jacob PhelpsJacob Phelps
161
$endgroup$
Solved 2 x 6 x (9!) = 4,354,560.
My error was using a 7 where the 6 was, forgot to consider the 2 at the beginning
answered Dec 11 '18 at 8:14
Jacob PhelpsJacob Phelps
161
answered Dec 11 '18 at 8:14
Jacob PhelpsJacob Phelps
161
answered Dec 11 '18 at 8:14
Jacob PhelpsJacob Phelps
161
answered Dec 11 '18 at 8:14
Jacob PhelpsJacob Phelps
161
161
1
$begingroup$
Your answer is incorrect. You have not explained how you got your answer, so I am not sure what you did. Since the two $I$s are indistinguishable, you should have divided by $2$.
$endgroup$
– N. F. Taussig
Dec 11 '18 at 9:36
$begingroup$
Welcome to MathSE. The expectation at this site is that you include your attempt in the question itself, which you can do by clicking on the edit button at the bottom of your question. Explaining how you arrived at your answer helps users of this site identify any errors you may have made. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Dec 11 '18 at 9:40
add a comment |
1
$begingroup$
Your answer is incorrect. You have not explained how you got your answer, so I am not sure what you did. Since the two $I$s are indistinguishable, you should have divided by $2$.
$endgroup$
– N. F. Taussig
Dec 11 '18 at 9:36
$begingroup$
Welcome to MathSE. The expectation at this site is that you include your attempt in the question itself, which you can do by clicking on the edit button at the bottom of your question. Explaining how you arrived at your answer helps users of this site identify any errors you may have made. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Dec 11 '18 at 9:40
1
1
$begingroup$
Your answer is incorrect. You have not explained how you got your answer, so I am not sure what you did. Since the two $I$s are indistinguishable, you should have divided by $2$.
$endgroup$
– N. F. Taussig
Dec 11 '18 at 9:36
$begingroup$
Your answer is incorrect. You have not explained how you got your answer, so I am not sure what you did. Since the two $I$s are indistinguishable, you should have divided by $2$.
$endgroup$
– N. F. Taussig
Dec 11 '18 at 9:36
$begingroup$
Welcome to MathSE. The expectation at this site is that you include your attempt in the question itself, which you can do by clicking on the edit button at the bottom of your question. Explaining how you arrived at your answer helps users of this site identify any errors you may have made. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Dec 11 '18 at 9:40
$begingroup$
Welcome to MathSE. The expectation at this site is that you include your attempt in the question itself, which you can do by clicking on the edit button at the bottom of your question. Explaining how you arrived at your answer helps users of this site identify any errors you may have made. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Dec 11 '18 at 9:40
add a comment |
