How many 5-letter words can we make if the letters are in order?How many two letter words can be formed from 26 English letters?K Permutations, 7 letter string, alphabet, no repeatsHow many $3$ letter words can be formed from the word “$TESTBOOK$”?Find the number of 3-letter words that can be made with letters in alphabetical order.Find the number of five-letter words that use letters from $A, B, C, ldots, Z$ in which all letters are different and are in alphabetical order.How many 4-letter words have all letters in alphabetical order?How many 10 letter words can be formed using a,b,c,d,e,f with the following conditionHow many 5 letter words can be formed from the word management if two alike letters are always togetherHow many distinct four-letter words beginning with A can be formed from letters with two similar letters and two different letters?How many n-letter words are there, such that number of letters “a” is even?
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How many 5-letter words can we make if the letters are in order?
How many two letter words can be formed from 26 English letters?K Permutations, 7 letter string, alphabet, no repeatsHow many $3$ letter words can be formed from the word “$TESTBOOK$”?Find the number of 3-letter words that can be made with letters in alphabetical order.Find the number of five-letter words that use letters from $A, B, C, ldots, Z$ in which all letters are different and are in alphabetical order.How many 4-letter words have all letters in alphabetical order?How many 10 letter words can be formed using a,b,c,d,e,f with the following conditionHow many 5 letter words can be formed from the word management if two alike letters are always togetherHow many distinct four-letter words beginning with A can be formed from letters with two similar letters and two different letters?How many n-letter words are there, such that number of letters “a” is even?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
Using the $26$ English letters, the number of $5$-letter words that can be made if the letters are distinct is determined as follows:
$26P5=26times25times24times23times22=7893600$ different words.
What if the letters in each word are in alphabetical order?
For example, the word JLOQY is valid, but the word JUMPY is invalid since U can not be before M
combinatorics statistics permutations combinations
asked Jul 6 at 11:38
Hussain-AlqatariHussain-Alqatari
6647 bronze badges
$endgroup$
add a comment |
$begingroup$
Using the $26$ English letters, the number of $5$-letter words that can be made if the letters are distinct is determined as follows:
$26P5=26times25times24times23times22=7893600$ different words.
What if the letters in each word are in alphabetical order?
For example, the word JLOQY is valid, but the word JUMPY is invalid since U can not be before M
combinatorics statistics permutations combinations
asked Jul 6 at 11:38
Hussain-AlqatariHussain-Alqatari
6647 bronze badges
$endgroup$
$begingroup$
Do you allow repeats for the main question (e.g. ABBEY)?
$endgroup$
– Parcly Taxel
Jul 6 at 11:42
$begingroup$
@ParclyTaxel Repetition is not allowed ,,, like when calculating 26P5.
$endgroup$
– Hussain-Alqatari
Jul 6 at 11:44
$begingroup$
As an aside, I prefer to think of the result as a binomial coefficient $binom265$ rather than as a falling factorial divided by a factorial $frac26frac5~5!$, or using your notation $frac~_26P_55!$
$endgroup$
– JMoravitz
Jul 6 at 11:56
add a comment |
$begingroup$
Using the $26$ English letters, the number of $5$-letter words that can be made if the letters are distinct is determined as follows:
$26P5=26times25times24times23times22=7893600$ different words.
What if the letters in each word are in alphabetical order?
For example, the word JLOQY is valid, but the word JUMPY is invalid since U can not be before M
combinatorics statistics permutations combinations
asked Jul 6 at 11:38
Hussain-AlqatariHussain-Alqatari
6647 bronze badges
$endgroup$
Using the $26$ English letters, the number of $5$-letter words that can be made if the letters are distinct is determined as follows:
$26P5=26times25times24times23times22=7893600$ different words.
What if the letters in each word are in alphabetical order?
For example, the word JLOQY is valid, but the word JUMPY is invalid since U can not be before M
combinatorics statistics permutations combinations
combinatorics statistics permutations combinations
asked Jul 6 at 11:38
Hussain-AlqatariHussain-Alqatari
6647 bronze badges
asked Jul 6 at 11:38
Hussain-AlqatariHussain-Alqatari
6647 bronze badges
asked Jul 6 at 11:38
Hussain-AlqatariHussain-Alqatari
6647 bronze badges
asked Jul 6 at 11:38
Hussain-AlqatariHussain-Alqatari
6647 bronze badges
asked Jul 6 at 11:38
Hussain-AlqatariHussain-Alqatari
6647 bronze badges
6647 bronze badges
$begingroup$
Do you allow repeats for the main question (e.g. ABBEY)?
$endgroup$
– Parcly Taxel
Jul 6 at 11:42
$begingroup$
@ParclyTaxel Repetition is not allowed ,,, like when calculating 26P5.
$endgroup$
– Hussain-Alqatari
Jul 6 at 11:44
$begingroup$
As an aside, I prefer to think of the result as a binomial coefficient $binom265$ rather than as a falling factorial divided by a factorial $frac26frac5~5!$, or using your notation $frac~_26P_55!$
$endgroup$
– JMoravitz
Jul 6 at 11:56
add a comment |
$begingroup$
Do you allow repeats for the main question (e.g. ABBEY)?
$endgroup$
– Parcly Taxel
Jul 6 at 11:42
$begingroup$
@ParclyTaxel Repetition is not allowed ,,, like when calculating 26P5.
$endgroup$
– Hussain-Alqatari
Jul 6 at 11:44
$begingroup$
As an aside, I prefer to think of the result as a binomial coefficient $binom265$ rather than as a falling factorial divided by a factorial $frac26frac5~5!$, or using your notation $frac~_26P_55!$
$endgroup$
– JMoravitz
Jul 6 at 11:56
$begingroup$
Do you allow repeats for the main question (e.g. ABBEY)?
$endgroup$
– Parcly Taxel
Jul 6 at 11:42
$begingroup$
Do you allow repeats for the main question (e.g. ABBEY)?
$endgroup$
– Parcly Taxel
Jul 6 at 11:42
$begingroup$
@ParclyTaxel Repetition is not allowed ,,, like when calculating 26P5.
$endgroup$
– Hussain-Alqatari
Jul 6 at 11:44
$begingroup$
@ParclyTaxel Repetition is not allowed ,,, like when calculating 26P5.
$endgroup$
– Hussain-Alqatari
Jul 6 at 11:44
$begingroup$
As an aside, I prefer to think of the result as a binomial coefficient $binom265$ rather than as a falling factorial divided by a factorial $frac26frac5~5!$, or using your notation $frac~_26P_55!$
$endgroup$
– JMoravitz
Jul 6 at 11:56
$begingroup$
As an aside, I prefer to think of the result as a binomial coefficient $binom265$ rather than as a falling factorial divided by a factorial $frac26frac5~5!$, or using your notation $frac~_26P_55!$
$endgroup$
– JMoravitz
Jul 6 at 11:56
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint. How many ways can you choose the five different letters? Once you have them, now many ways can you organize them in alphabetical order?
(This assumes the letters are distinct.)
answered Jul 6 at 11:43
Ethan BolkerEthan Bolker
52.6k5 gold badges61 silver badges131 bronze badges
$endgroup$
1
$begingroup$
So will it be $frac26P55!=frac7893600120=65780$ different words?
$endgroup$
– Hussain-Alqatari
Jul 6 at 11:46
$begingroup$
Yes, that's right.
$endgroup$
– Ethan Bolker
Jul 6 at 11:54
add a comment |
$begingroup$
Divide out the number of permutations of five letters ($5!$), since only one is correct.
answered Jul 6 at 11:49
Chris CusterChris Custer
17.5k3 gold badges8 silver badges31 bronze badges
$endgroup$
add a comment |
$begingroup$
The different number of ways to select 5 alphabets from 26 alphabets= $26C5$.
Arrange the alphabets each collection in the required order.
Thus the answer is $26C5$.
answered Jul 6 at 11:57
Avinash NAvinash N
7594 silver badges11 bronze badges
$endgroup$
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
$begingroup$
Hint. How many ways can you choose the five different letters? Once you have them, now many ways can you organize them in alphabetical order?
(This assumes the letters are distinct.)
answered Jul 6 at 11:43
Ethan BolkerEthan Bolker
52.6k5 gold badges61 silver badges131 bronze badges
$endgroup$
1
$begingroup$
So will it be $frac26P55!=frac7893600120=65780$ different words?
$endgroup$
– Hussain-Alqatari
Jul 6 at 11:46
$begingroup$
Yes, that's right.
$endgroup$
– Ethan Bolker
Jul 6 at 11:54
add a comment |
$begingroup$
Hint. How many ways can you choose the five different letters? Once you have them, now many ways can you organize them in alphabetical order?
(This assumes the letters are distinct.)
answered Jul 6 at 11:43
Ethan BolkerEthan Bolker
52.6k5 gold badges61 silver badges131 bronze badges
$endgroup$
1
$begingroup$
So will it be $frac26P55!=frac7893600120=65780$ different words?
$endgroup$
– Hussain-Alqatari
Jul 6 at 11:46
$begingroup$
Yes, that's right.
$endgroup$
– Ethan Bolker
Jul 6 at 11:54
add a comment |
$begingroup$
Hint. How many ways can you choose the five different letters? Once you have them, now many ways can you organize them in alphabetical order?
(This assumes the letters are distinct.)
answered Jul 6 at 11:43
Ethan BolkerEthan Bolker
52.6k5 gold badges61 silver badges131 bronze badges
$endgroup$
Hint. How many ways can you choose the five different letters? Once you have them, now many ways can you organize them in alphabetical order?
(This assumes the letters are distinct.)
answered Jul 6 at 11:43
Ethan BolkerEthan Bolker
52.6k5 gold badges61 silver badges131 bronze badges
answered Jul 6 at 11:43
Ethan BolkerEthan Bolker
52.6k5 gold badges61 silver badges131 bronze badges
answered Jul 6 at 11:43
Ethan BolkerEthan Bolker
52.6k5 gold badges61 silver badges131 bronze badges
answered Jul 6 at 11:43
Ethan BolkerEthan Bolker
52.6k5 gold badges61 silver badges131 bronze badges
52.6k5 gold badges61 silver badges131 bronze badges
1
$begingroup$
So will it be $frac26P55!=frac7893600120=65780$ different words?
$endgroup$
– Hussain-Alqatari
Jul 6 at 11:46
$begingroup$
Yes, that's right.
$endgroup$
– Ethan Bolker
Jul 6 at 11:54
add a comment |
1
$begingroup$
So will it be $frac26P55!=frac7893600120=65780$ different words?
$endgroup$
– Hussain-Alqatari
Jul 6 at 11:46
$begingroup$
Yes, that's right.
$endgroup$
– Ethan Bolker
Jul 6 at 11:54
1
1
$begingroup$
So will it be $frac26P55!=frac7893600120=65780$ different words?
$endgroup$
– Hussain-Alqatari
Jul 6 at 11:46
$begingroup$
So will it be $frac26P55!=frac7893600120=65780$ different words?
$endgroup$
– Hussain-Alqatari
Jul 6 at 11:46
$begingroup$
Yes, that's right.
$endgroup$
– Ethan Bolker
Jul 6 at 11:54
$begingroup$
Yes, that's right.
$endgroup$
– Ethan Bolker
Jul 6 at 11:54
add a comment |
$begingroup$
Divide out the number of permutations of five letters ($5!$), since only one is correct.
answered Jul 6 at 11:49
Chris CusterChris Custer
17.5k3 gold badges8 silver badges31 bronze badges
$endgroup$
add a comment |
$begingroup$
Divide out the number of permutations of five letters ($5!$), since only one is correct.
answered Jul 6 at 11:49
Chris CusterChris Custer
17.5k3 gold badges8 silver badges31 bronze badges
$endgroup$
add a comment |
$begingroup$
Divide out the number of permutations of five letters ($5!$), since only one is correct.
answered Jul 6 at 11:49
Chris CusterChris Custer
17.5k3 gold badges8 silver badges31 bronze badges
$endgroup$
Divide out the number of permutations of five letters ($5!$), since only one is correct.
answered Jul 6 at 11:49
Chris CusterChris Custer
17.5k3 gold badges8 silver badges31 bronze badges
answered Jul 6 at 11:49
Chris CusterChris Custer
17.5k3 gold badges8 silver badges31 bronze badges
answered Jul 6 at 11:49
Chris CusterChris Custer
17.5k3 gold badges8 silver badges31 bronze badges
answered Jul 6 at 11:49
Chris CusterChris Custer
17.5k3 gold badges8 silver badges31 bronze badges
17.5k3 gold badges8 silver badges31 bronze badges
add a comment |
add a comment |
$begingroup$
The different number of ways to select 5 alphabets from 26 alphabets= $26C5$.
Arrange the alphabets each collection in the required order.
Thus the answer is $26C5$.
answered Jul 6 at 11:57
Avinash NAvinash N
7594 silver badges11 bronze badges
$endgroup$
add a comment |
$begingroup$
The different number of ways to select 5 alphabets from 26 alphabets= $26C5$.
Arrange the alphabets each collection in the required order.
Thus the answer is $26C5$.
answered Jul 6 at 11:57
Avinash NAvinash N
7594 silver badges11 bronze badges
$endgroup$
add a comment |
$begingroup$
The different number of ways to select 5 alphabets from 26 alphabets= $26C5$.
Arrange the alphabets each collection in the required order.
Thus the answer is $26C5$.
answered Jul 6 at 11:57
Avinash NAvinash N
7594 silver badges11 bronze badges
$endgroup$
The different number of ways to select 5 alphabets from 26 alphabets= $26C5$.
Arrange the alphabets each collection in the required order.
Thus the answer is $26C5$.
answered Jul 6 at 11:57
Avinash NAvinash N
7594 silver badges11 bronze badges
answered Jul 6 at 11:57
Avinash NAvinash N
7594 silver badges11 bronze badges
answered Jul 6 at 11:57
Avinash NAvinash N
7594 silver badges11 bronze badges
answered Jul 6 at 11:57
Avinash NAvinash N
7594 silver badges11 bronze badges
7594 silver badges11 bronze badges
add a comment |
add a comment |
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$begingroup$
Do you allow repeats for the main question (e.g. ABBEY)?
$endgroup$
– Parcly Taxel
Jul 6 at 11:42
$begingroup$
@ParclyTaxel Repetition is not allowed ,,, like when calculating 26P5.
$endgroup$
– Hussain-Alqatari
Jul 6 at 11:44
$begingroup$
As an aside, I prefer to think of the result as a binomial coefficient $binom265$ rather than as a falling factorial divided by a factorial $frac26frac5~5!$, or using your notation $frac~_26P_55!$
$endgroup$
– JMoravitz
Jul 6 at 11:56