How many of them will be palindromes give the following conditions?In how many ways can one or more of $5$...
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How many of them will be palindromes give the following conditions?
In how many ways can one or more of $5$ letters be posted into $4$ mail boxes, if any letter can be posted into any of the boxes?how to arrange vowels be arranged in even places given the following conditions?In how many ways can the letters of english alphabets is arranged given the following conditions?Combination and Permutation How many words can be formed?How many $3$ letter words can be formed from the word “$TESTBOOK$”?Number of Four letter wordsHow many words can be formed from all the letters of the word 'INITIAL' that start and end with the letter 'I'?Number of palindromes in the alphabetHow many $10$-letter words can be formed using the $26$ letters of the alphabet if repetition is allowed, but letters are listed alphabeticallyIn how many ways can the letters in the word “PROBABILITY” be arranged using the following restrcitions
$begingroup$
A 5 letter word is formed using some of the letters (a,b,h,i,p,r,s}
How many of them will be palindromes?
Options
a)$125$ b)$225$ c)$343$ d)$729$
My Approach:
Only the middle number can be arranged in $7$ ways.Rest all the numbers can be arranged like Whatever the first letter from right should be last letter from left.Whatever the second letter from start should be last letter from left.
But i am not able to get to the solution.
Can anyone give me the hint to solve the problem.
permutations
asked Sep 27 '15 at 11:02
justin takrojustin takro
69031835
$endgroup$
add a comment |
$begingroup$
A 5 letter word is formed using some of the letters (a,b,h,i,p,r,s}
How many of them will be palindromes?
Options
a)$125$ b)$225$ c)$343$ d)$729$
My Approach:
Only the middle number can be arranged in $7$ ways.Rest all the numbers can be arranged like Whatever the first letter from right should be last letter from left.Whatever the second letter from start should be last letter from left.
But i am not able to get to the solution.
Can anyone give me the hint to solve the problem.
permutations
asked Sep 27 '15 at 11:02
justin takrojustin takro
69031835
$endgroup$
add a comment |
$begingroup$
A 5 letter word is formed using some of the letters (a,b,h,i,p,r,s}
How many of them will be palindromes?
Options
a)$125$ b)$225$ c)$343$ d)$729$
My Approach:
Only the middle number can be arranged in $7$ ways.Rest all the numbers can be arranged like Whatever the first letter from right should be last letter from left.Whatever the second letter from start should be last letter from left.
But i am not able to get to the solution.
Can anyone give me the hint to solve the problem.
permutations
asked Sep 27 '15 at 11:02
justin takrojustin takro
69031835
$endgroup$
A 5 letter word is formed using some of the letters (a,b,h,i,p,r,s}
How many of them will be palindromes?
Options
a)$125$ b)$225$ c)$343$ d)$729$
My Approach:
Only the middle number can be arranged in $7$ ways.Rest all the numbers can be arranged like Whatever the first letter from right should be last letter from left.Whatever the second letter from start should be last letter from left.
But i am not able to get to the solution.
Can anyone give me the hint to solve the problem.
permutations
permutations
asked Sep 27 '15 at 11:02
justin takrojustin takro
69031835
asked Sep 27 '15 at 11:02
justin takrojustin takro
69031835
asked Sep 27 '15 at 11:02
justin takrojustin takro
69031835
asked Sep 27 '15 at 11:02
justin takrojustin takro
69031835
asked Sep 27 '15 at 11:02
justin takrojustin takro
69031835
69031835
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
**HINT:**Think about how many ways you have to choose the first letter, then the second, then the third and so on...
answered Sep 27 '15 at 11:45
karmalukarmalu
897310
$endgroup$
add a comment |
$begingroup$
Pick one of the $7$ letters as the middle letter, another one as the second and fourth letter, and finally one more for the first and last letter.
Assuming repetitions are allowed, there are $7cdot7cdot7$ ways to do this.
answered Sep 27 '15 at 11:45
Yiyuan LeeYiyuan Lee
12.9k42960
$endgroup$
add a comment |
$begingroup$
Pick the first center letter. At the right side you have a permutation of 7 letters in two i.e you 42 left and right letters for each center letter. So you will have 42x7 = 294 then add aa, bb, hh, ii,...for each center letter you will have 49 more. So the answer is c.
answered Mar 19 at 1:25
user655572user655572
1
$endgroup$
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:**Think about how many ways you have to choose the first letter, then the second, then the third and so on...
answered Sep 27 '15 at 11:45
karmalukarmalu
897310
$endgroup$
add a comment |
$begingroup$
**HINT:**Think about how many ways you have to choose the first letter, then the second, then the third and so on...
answered Sep 27 '15 at 11:45
karmalukarmalu
897310
$endgroup$
add a comment |
$begingroup$
**HINT:**Think about how many ways you have to choose the first letter, then the second, then the third and so on...
answered Sep 27 '15 at 11:45
karmalukarmalu
897310
$endgroup$
**HINT:**Think about how many ways you have to choose the first letter, then the second, then the third and so on...
answered Sep 27 '15 at 11:45
karmalukarmalu
897310
answered Sep 27 '15 at 11:45
karmalukarmalu
897310
answered Sep 27 '15 at 11:45
karmalukarmalu
897310
answered Sep 27 '15 at 11:45
karmalukarmalu
897310
897310
add a comment |
add a comment |
$begingroup$
Pick one of the $7$ letters as the middle letter, another one as the second and fourth letter, and finally one more for the first and last letter.
Assuming repetitions are allowed, there are $7cdot7cdot7$ ways to do this.
answered Sep 27 '15 at 11:45
Yiyuan LeeYiyuan Lee
12.9k42960
$endgroup$
add a comment |
$begingroup$
Pick one of the $7$ letters as the middle letter, another one as the second and fourth letter, and finally one more for the first and last letter.
Assuming repetitions are allowed, there are $7cdot7cdot7$ ways to do this.
answered Sep 27 '15 at 11:45
Yiyuan LeeYiyuan Lee
12.9k42960
$endgroup$
add a comment |
$begingroup$
Pick one of the $7$ letters as the middle letter, another one as the second and fourth letter, and finally one more for the first and last letter.
Assuming repetitions are allowed, there are $7cdot7cdot7$ ways to do this.
answered Sep 27 '15 at 11:45
Yiyuan LeeYiyuan Lee
12.9k42960
$endgroup$
Pick one of the $7$ letters as the middle letter, another one as the second and fourth letter, and finally one more for the first and last letter.
Assuming repetitions are allowed, there are $7cdot7cdot7$ ways to do this.
answered Sep 27 '15 at 11:45
Yiyuan LeeYiyuan Lee
12.9k42960
answered Sep 27 '15 at 11:45
Yiyuan LeeYiyuan Lee
12.9k42960
answered Sep 27 '15 at 11:45
Yiyuan LeeYiyuan Lee
12.9k42960
answered Sep 27 '15 at 11:45
Yiyuan LeeYiyuan Lee
12.9k42960
12.9k42960
add a comment |
add a comment |
$begingroup$
Pick the first center letter. At the right side you have a permutation of 7 letters in two i.e you 42 left and right letters for each center letter. So you will have 42x7 = 294 then add aa, bb, hh, ii,...for each center letter you will have 49 more. So the answer is c.
answered Mar 19 at 1:25
user655572user655572
1
$endgroup$
add a comment |
$begingroup$
Pick the first center letter. At the right side you have a permutation of 7 letters in two i.e you 42 left and right letters for each center letter. So you will have 42x7 = 294 then add aa, bb, hh, ii,...for each center letter you will have 49 more. So the answer is c.
answered Mar 19 at 1:25
user655572user655572
1
$endgroup$
add a comment |
$begingroup$
Pick the first center letter. At the right side you have a permutation of 7 letters in two i.e you 42 left and right letters for each center letter. So you will have 42x7 = 294 then add aa, bb, hh, ii,...for each center letter you will have 49 more. So the answer is c.
answered Mar 19 at 1:25
user655572user655572
1
$endgroup$
Pick the first center letter. At the right side you have a permutation of 7 letters in two i.e you 42 left and right letters for each center letter. So you will have 42x7 = 294 then add aa, bb, hh, ii,...for each center letter you will have 49 more. So the answer is c.
answered Mar 19 at 1:25
user655572user655572
1
answered Mar 19 at 1:25
user655572user655572
1
answered Mar 19 at 1:25
user655572user655572
1
answered Mar 19 at 1:25
user655572user655572
1
1
add a comment |
add a comment |
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