Number of words that can be made using all the letters of the word W, if Os as well as Is are separated is? The Next CEO of Stack OverflowNumber of rearrangements of the word “INDIVISIBILITY”How many words, with or without meaning, each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE?How many different words can be formed with the letters of 'SKYSCRAPERS' if A and E are together?Forming seven letter words by using the letters of the word $textSUCCESS$Total no. of words formed when similar letters are involvedNumber of ways to arrange $A,A,A,B,C,C$ such that no $2$ consecutive letters are the sameFind the total number of permutations in which letters of the word $ARRANGEMENT$ can be permuted so that two $E$s and two $R$s do not come together.Find the number of ways in which 6 boys and 6 girls can be seated in a row so that all the girls are never together.What is the number of arrangements in the word “EDUCATION” where vowels are never together?Probability that the first 2 letters are consonants when the letters of the word 'equilibrium' are rearranged
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Number of words that can be made using all the letters of the word W, if Os as well as Is are separated is?
The Next CEO of Stack OverflowNumber of rearrangements of the word “INDIVISIBILITY”How many words, with or without meaning, each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE?How many different words can be formed with the letters of 'SKYSCRAPERS' if A and E are together?Forming seven letter words by using the letters of the word $textSUCCESS$Total no. of words formed when similar letters are involvedNumber of ways to arrange $A,A,A,B,C,C$ such that no $2$ consecutive letters are the sameFind the total number of permutations in which letters of the word $ARRANGEMENT$ can be permuted so that two $E$s and two $R$s do not come together.Find the number of ways in which 6 boys and 6 girls can be seated in a row so that all the girls are never together.What is the number of arrangements in the word “EDUCATION” where vowels are never together?Probability that the first 2 letters are consonants when the letters of the word 'equilibrium' are rearranged
$begingroup$
I am facing difficulty in solving the 18th Question from the above passage.
Attempt:
I have attempted it using principal of inclusion and exclusion.
Let n be the required number of ways.
$n = textTotal number of ways - Number of ways in which Os and Is are together$
$implies n = dfrac12!3!2!2! - dfrac11!2!3! - dfrac10!2!2!+ dfrac9!2!2! = 399 times 8!$
This gives $N = 399$ which is wrong. Can you please tell me why am I getting the wrong answer?
combinatorics permutations combinations
asked 23 hours ago
AbcdAbcd
3,10931338
$endgroup$
add a comment |
$begingroup$
I am facing difficulty in solving the 18th Question from the above passage.
Attempt:
I have attempted it using principal of inclusion and exclusion.
Let n be the required number of ways.
$n = textTotal number of ways - Number of ways in which Os and Is are together$
$implies n = dfrac12!3!2!2! - dfrac11!2!3! - dfrac10!2!2!+ dfrac9!2!2! = 399 times 8!$
This gives $N = 399$ which is wrong. Can you please tell me why am I getting the wrong answer?
combinatorics permutations combinations
asked 23 hours ago
AbcdAbcd
3,10931338
$endgroup$
1
$begingroup$
How do you know that your solution is wrong? There is also option (D)
$endgroup$
– Robert Z
21 hours ago
$begingroup$
@RobertZ answer given is option B
$endgroup$
– Abcd
21 hours ago
add a comment |
$begingroup$
I am facing difficulty in solving the 18th Question from the above passage.
Attempt:
I have attempted it using principal of inclusion and exclusion.
Let n be the required number of ways.
$n = textTotal number of ways - Number of ways in which Os and Is are together$
$implies n = dfrac12!3!2!2! - dfrac11!2!3! - dfrac10!2!2!+ dfrac9!2!2! = 399 times 8!$
This gives $N = 399$ which is wrong. Can you please tell me why am I getting the wrong answer?
combinatorics permutations combinations
asked 23 hours ago
AbcdAbcd
3,10931338
$endgroup$
I am facing difficulty in solving the 18th Question from the above passage.
Attempt:
I have attempted it using principal of inclusion and exclusion.
Let n be the required number of ways.
$n = textTotal number of ways - Number of ways in which Os and Is are together$
$implies n = dfrac12!3!2!2! - dfrac11!2!3! - dfrac10!2!2!+ dfrac9!2!2! = 399 times 8!$
This gives $N = 399$ which is wrong. Can you please tell me why am I getting the wrong answer?
combinatorics permutations combinations
combinatorics permutations combinations
asked 23 hours ago
AbcdAbcd
3,10931338
asked 23 hours ago
AbcdAbcd
3,10931338
asked 23 hours ago
AbcdAbcd
3,10931338
asked 23 hours ago
AbcdAbcd
3,10931338
asked 23 hours ago
AbcdAbcd
3,10931338
3,10931338
1
$begingroup$
How do you know that your solution is wrong? There is also option (D)
$endgroup$
– Robert Z
21 hours ago
$begingroup$
@RobertZ answer given is option B
$endgroup$
– Abcd
21 hours ago
add a comment |
1
$begingroup$
How do you know that your solution is wrong? There is also option (D)
$endgroup$
– Robert Z
21 hours ago
$begingroup$
@RobertZ answer given is option B
$endgroup$
– Abcd
21 hours ago
1
1
$begingroup$
How do you know that your solution is wrong? There is also option (D)
$endgroup$
– Robert Z
21 hours ago
$begingroup$
How do you know that your solution is wrong? There is also option (D)
$endgroup$
– Robert Z
21 hours ago
$begingroup$
@RobertZ answer given is option B
$endgroup$
– Abcd
21 hours ago
$begingroup$
@RobertZ answer given is option B
$endgroup$
– Abcd
21 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Your mistake is not noticing that the I's can be non-separated two different ways: We can have all three of them together as III, but we can also have II in one place and I in another.
One way to calculate how many ways this can happen is to think of one pair II as a single letter, and the third I as a separate letter, and count the number of words we can make. This will count each III case twice: Once as I II, and once as II I. So we have to subtract the number of such cases once.
With this correction, the final calculation becomes
$$
overbracefrac12!3!2!2!^textAll words - overbracefrac11!3!2!^textO's not separate - overbraceleft(underbracefrac11!2!2!_textII and I - underbracefrac10!2!2!_textIIIright)^textI's not separate + overbraceleft(underbracefrac10!2!_textOO, II and I - underbracefrac9!2!_textOO and IIIright)^textNeither O's nor I's separate
$$
which turns out to be $8!cdot 228$.
answered 21 hours ago
ArthurArthur
120k7121206
$endgroup$
$begingroup$
Thanks for the answer I have understood my mistake, but I am unable to understand the final equation in your answer. Can you label the terms like which term stands for which case?
$endgroup$
– Abcd
21 hours ago
$begingroup$
@Abcd I added a short explanation for what each term represents. The basic structure of your inclusion-exclusion calculation is still there, just with the two last terms corrected to make up for the mistake.
$endgroup$
– Arthur
20 hours ago
1
$begingroup$
+1 for the MathJax magic!
$endgroup$
– TonyK
18 hours ago
add a comment |
$begingroup$
The letter $I$ appears three times. You have calculated the number of ways to arrange the letters so that the three copies of $I$ aren't all in the same place (and also the two $O$s are separated). However, "separated" here means that no two of them are together. Your count includes such "words" as STIILOCATINO, which should be excluded.
answered 21 hours ago
jmerryjmerry
16.8k11633
$endgroup$
add a comment |
$begingroup$
How did you get each of the terms of $n$?
$frac12!3!2!2!$ : The number of ways of arranging the letters of SOLICITATION.
$frac11!2!3!$ : The number of ways in which the two $O$s can be together in SOLICITATION.
$frac10!2!2!$ : The number of ways in which the three $I$s can be together in SOLICITATION.
$frac9!2!2!$ : The number of ways in which the three $I$s and two $O$s can be together in SOLICITATION.
Well , the problem is this : the complement of this case, also includes the case where two $I$s can be together and the third can be separate. You have not neglected this case, and that's why you have an answer larger than the given answer.
answered 21 hours ago
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40k33477
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your mistake is not noticing that the I's can be non-separated two different ways: We can have all three of them together as III, but we can also have II in one place and I in another.
One way to calculate how many ways this can happen is to think of one pair II as a single letter, and the third I as a separate letter, and count the number of words we can make. This will count each III case twice: Once as I II, and once as II I. So we have to subtract the number of such cases once.
With this correction, the final calculation becomes
$$
overbracefrac12!3!2!2!^textAll words - overbracefrac11!3!2!^textO's not separate - overbraceleft(underbracefrac11!2!2!_textII and I - underbracefrac10!2!2!_textIIIright)^textI's not separate + overbraceleft(underbracefrac10!2!_textOO, II and I - underbracefrac9!2!_textOO and IIIright)^textNeither O's nor I's separate
$$
which turns out to be $8!cdot 228$.
answered 21 hours ago
ArthurArthur
120k7121206
$endgroup$
$begingroup$
Thanks for the answer I have understood my mistake, but I am unable to understand the final equation in your answer. Can you label the terms like which term stands for which case?
$endgroup$
– Abcd
21 hours ago
$begingroup$
@Abcd I added a short explanation for what each term represents. The basic structure of your inclusion-exclusion calculation is still there, just with the two last terms corrected to make up for the mistake.
$endgroup$
– Arthur
20 hours ago
1
$begingroup$
+1 for the MathJax magic!
$endgroup$
– TonyK
18 hours ago
add a comment |
$begingroup$
Your mistake is not noticing that the I's can be non-separated two different ways: We can have all three of them together as III, but we can also have II in one place and I in another.
One way to calculate how many ways this can happen is to think of one pair II as a single letter, and the third I as a separate letter, and count the number of words we can make. This will count each III case twice: Once as I II, and once as II I. So we have to subtract the number of such cases once.
With this correction, the final calculation becomes
$$
overbracefrac12!3!2!2!^textAll words - overbracefrac11!3!2!^textO's not separate - overbraceleft(underbracefrac11!2!2!_textII and I - underbracefrac10!2!2!_textIIIright)^textI's not separate + overbraceleft(underbracefrac10!2!_textOO, II and I - underbracefrac9!2!_textOO and IIIright)^textNeither O's nor I's separate
$$
which turns out to be $8!cdot 228$.
answered 21 hours ago
ArthurArthur
120k7121206
$endgroup$
$begingroup$
Thanks for the answer I have understood my mistake, but I am unable to understand the final equation in your answer. Can you label the terms like which term stands for which case?
$endgroup$
– Abcd
21 hours ago
$begingroup$
@Abcd I added a short explanation for what each term represents. The basic structure of your inclusion-exclusion calculation is still there, just with the two last terms corrected to make up for the mistake.
$endgroup$
– Arthur
20 hours ago
1
$begingroup$
+1 for the MathJax magic!
$endgroup$
– TonyK
18 hours ago
add a comment |
$begingroup$
Your mistake is not noticing that the I's can be non-separated two different ways: We can have all three of them together as III, but we can also have II in one place and I in another.
One way to calculate how many ways this can happen is to think of one pair II as a single letter, and the third I as a separate letter, and count the number of words we can make. This will count each III case twice: Once as I II, and once as II I. So we have to subtract the number of such cases once.
With this correction, the final calculation becomes
$$
overbracefrac12!3!2!2!^textAll words - overbracefrac11!3!2!^textO's not separate - overbraceleft(underbracefrac11!2!2!_textII and I - underbracefrac10!2!2!_textIIIright)^textI's not separate + overbraceleft(underbracefrac10!2!_textOO, II and I - underbracefrac9!2!_textOO and IIIright)^textNeither O's nor I's separate
$$
which turns out to be $8!cdot 228$.
answered 21 hours ago
ArthurArthur
120k7121206
$endgroup$
Your mistake is not noticing that the I's can be non-separated two different ways: We can have all three of them together as III, but we can also have II in one place and I in another.
One way to calculate how many ways this can happen is to think of one pair II as a single letter, and the third I as a separate letter, and count the number of words we can make. This will count each III case twice: Once as I II, and once as II I. So we have to subtract the number of such cases once.
With this correction, the final calculation becomes
$$
overbracefrac12!3!2!2!^textAll words - overbracefrac11!3!2!^textO's not separate - overbraceleft(underbracefrac11!2!2!_textII and I - underbracefrac10!2!2!_textIIIright)^textI's not separate + overbraceleft(underbracefrac10!2!_textOO, II and I - underbracefrac9!2!_textOO and IIIright)^textNeither O's nor I's separate
$$
which turns out to be $8!cdot 228$.
answered 21 hours ago
ArthurArthur
120k7121206
edited 20 hours ago
answered 21 hours ago
ArthurArthur
120k7121206
answered 21 hours ago
ArthurArthur
120k7121206
answered 21 hours ago
ArthurArthur
120k7121206
120k7121206
$begingroup$
Thanks for the answer I have understood my mistake, but I am unable to understand the final equation in your answer. Can you label the terms like which term stands for which case?
$endgroup$
– Abcd
21 hours ago
$begingroup$
@Abcd I added a short explanation for what each term represents. The basic structure of your inclusion-exclusion calculation is still there, just with the two last terms corrected to make up for the mistake.
$endgroup$
– Arthur
20 hours ago
1
$begingroup$
+1 for the MathJax magic!
$endgroup$
– TonyK
18 hours ago
add a comment |
$begingroup$
Thanks for the answer I have understood my mistake, but I am unable to understand the final equation in your answer. Can you label the terms like which term stands for which case?
$endgroup$
– Abcd
21 hours ago
$begingroup$
@Abcd I added a short explanation for what each term represents. The basic structure of your inclusion-exclusion calculation is still there, just with the two last terms corrected to make up for the mistake.
$endgroup$
– Arthur
20 hours ago
1
$begingroup$
+1 for the MathJax magic!
$endgroup$
– TonyK
18 hours ago
$begingroup$
Thanks for the answer I have understood my mistake, but I am unable to understand the final equation in your answer. Can you label the terms like which term stands for which case?
$endgroup$
– Abcd
21 hours ago
$begingroup$
Thanks for the answer I have understood my mistake, but I am unable to understand the final equation in your answer. Can you label the terms like which term stands for which case?
$endgroup$
– Abcd
21 hours ago
$begingroup$
@Abcd I added a short explanation for what each term represents. The basic structure of your inclusion-exclusion calculation is still there, just with the two last terms corrected to make up for the mistake.
$endgroup$
– Arthur
20 hours ago
$begingroup$
@Abcd I added a short explanation for what each term represents. The basic structure of your inclusion-exclusion calculation is still there, just with the two last terms corrected to make up for the mistake.
$endgroup$
– Arthur
20 hours ago
1
1
$begingroup$
+1 for the MathJax magic!
$endgroup$
– TonyK
18 hours ago
$begingroup$
+1 for the MathJax magic!
$endgroup$
– TonyK
18 hours ago
add a comment |
$begingroup$
The letter $I$ appears three times. You have calculated the number of ways to arrange the letters so that the three copies of $I$ aren't all in the same place (and also the two $O$s are separated). However, "separated" here means that no two of them are together. Your count includes such "words" as STIILOCATINO, which should be excluded.
answered 21 hours ago
jmerryjmerry
16.8k11633
$endgroup$
add a comment |
$begingroup$
The letter $I$ appears three times. You have calculated the number of ways to arrange the letters so that the three copies of $I$ aren't all in the same place (and also the two $O$s are separated). However, "separated" here means that no two of them are together. Your count includes such "words" as STIILOCATINO, which should be excluded.
answered 21 hours ago
jmerryjmerry
16.8k11633
$endgroup$
add a comment |
$begingroup$
The letter $I$ appears three times. You have calculated the number of ways to arrange the letters so that the three copies of $I$ aren't all in the same place (and also the two $O$s are separated). However, "separated" here means that no two of them are together. Your count includes such "words" as STIILOCATINO, which should be excluded.
answered 21 hours ago
jmerryjmerry
16.8k11633
$endgroup$
The letter $I$ appears three times. You have calculated the number of ways to arrange the letters so that the three copies of $I$ aren't all in the same place (and also the two $O$s are separated). However, "separated" here means that no two of them are together. Your count includes such "words" as STIILOCATINO, which should be excluded.
answered 21 hours ago
jmerryjmerry
16.8k11633
answered 21 hours ago
jmerryjmerry
16.8k11633
answered 21 hours ago
jmerryjmerry
16.8k11633
answered 21 hours ago
jmerryjmerry
16.8k11633
16.8k11633
add a comment |
add a comment |
$begingroup$
How did you get each of the terms of $n$?
$frac12!3!2!2!$ : The number of ways of arranging the letters of SOLICITATION.
$frac11!2!3!$ : The number of ways in which the two $O$s can be together in SOLICITATION.
$frac10!2!2!$ : The number of ways in which the three $I$s can be together in SOLICITATION.
$frac9!2!2!$ : The number of ways in which the three $I$s and two $O$s can be together in SOLICITATION.
Well , the problem is this : the complement of this case, also includes the case where two $I$s can be together and the third can be separate. You have not neglected this case, and that's why you have an answer larger than the given answer.
answered 21 hours ago
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40k33477
$endgroup$
add a comment |
$begingroup$
How did you get each of the terms of $n$?
$frac12!3!2!2!$ : The number of ways of arranging the letters of SOLICITATION.
$frac11!2!3!$ : The number of ways in which the two $O$s can be together in SOLICITATION.
$frac10!2!2!$ : The number of ways in which the three $I$s can be together in SOLICITATION.
$frac9!2!2!$ : The number of ways in which the three $I$s and two $O$s can be together in SOLICITATION.
Well , the problem is this : the complement of this case, also includes the case where two $I$s can be together and the third can be separate. You have not neglected this case, and that's why you have an answer larger than the given answer.
answered 21 hours ago
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40k33477
$endgroup$
add a comment |
$begingroup$
How did you get each of the terms of $n$?
$frac12!3!2!2!$ : The number of ways of arranging the letters of SOLICITATION.
$frac11!2!3!$ : The number of ways in which the two $O$s can be together in SOLICITATION.
$frac10!2!2!$ : The number of ways in which the three $I$s can be together in SOLICITATION.
$frac9!2!2!$ : The number of ways in which the three $I$s and two $O$s can be together in SOLICITATION.
Well , the problem is this : the complement of this case, also includes the case where two $I$s can be together and the third can be separate. You have not neglected this case, and that's why you have an answer larger than the given answer.
answered 21 hours ago
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40k33477
$endgroup$
How did you get each of the terms of $n$?
$frac12!3!2!2!$ : The number of ways of arranging the letters of SOLICITATION.
$frac11!2!3!$ : The number of ways in which the two $O$s can be together in SOLICITATION.
$frac10!2!2!$ : The number of ways in which the three $I$s can be together in SOLICITATION.
$frac9!2!2!$ : The number of ways in which the three $I$s and two $O$s can be together in SOLICITATION.
Well , the problem is this : the complement of this case, also includes the case where two $I$s can be together and the third can be separate. You have not neglected this case, and that's why you have an answer larger than the given answer.
answered 21 hours ago
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40k33477
answered 21 hours ago
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40k33477
answered 21 hours ago
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40k33477
answered 21 hours ago
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40k33477
40k33477
add a comment |
add a comment |
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1
$begingroup$
How do you know that your solution is wrong? There is also option (D)
$endgroup$
– Robert Z
21 hours ago
$begingroup$
@RobertZ answer given is option B
$endgroup$
– Abcd
21 hours ago