Permutation with no two vowels next to each other
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Question : find number of arrangements of the word TRIANGLE in which no two vowels are next to each other.
My attempt : $5! ( ^6P_3) =14,400$
Is this correct?
permutations
edited Aug 25 at 12:52
Key Flex
1
asked Aug 23 at 1:14
user122343
685
add a comment |Â
up vote
0
down vote
favorite
Question : find number of arrangements of the word TRIANGLE in which no two vowels are next to each other.
My attempt : $5! ( ^6P_3) =14,400$
Is this correct?
permutations
edited Aug 25 at 12:52
Key Flex
1
asked Aug 23 at 1:14
user122343
685
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Question : find number of arrangements of the word TRIANGLE in which no two vowels are next to each other.
My attempt : $5! ( ^6P_3) =14,400$
Is this correct?
permutations
edited Aug 25 at 12:52
Key Flex
1
asked Aug 23 at 1:14
user122343
685
Question : find number of arrangements of the word TRIANGLE in which no two vowels are next to each other.
My attempt : $5! ( ^6P_3) =14,400$
Is this correct?
permutations
edited Aug 25 at 12:52
Key Flex
1
asked Aug 23 at 1:14
user122343
685
edited Aug 25 at 12:52
Key Flex
1
edited Aug 25 at 12:52
Key Flex
1
edited Aug 25 at 12:52
Key Flex
1
1
asked Aug 23 at 1:14
user122343
685
asked Aug 23 at 1:14
user122343
685
asked Aug 23 at 1:14
user122343
685
685
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Yes, your answer is correct.
There are $5$ consonants and $3$ vowels.
Vowels can be selected in $dbinom63$ ways.
The vowels be arranged in $3!$ ways
The consonants can be arranged in $5!$ ways.
In total we have $dbinom63times3!times5!=14400$ ways.
answered Aug 23 at 1:18
Key Flex
1
1
Thank you . I will accept your answer with green tick :)
– user122343
Aug 23 at 1:21
@user122343 You are welcome :)
– Key Flex
Aug 23 at 1:22
Sorry I thought I did. I just did it back.
– user122343
Aug 25 at 6:01
add a comment |Â
up vote
0
down vote
The consonants can be arranged in $5!$ ways. There are then six "slots" available for single vowels: one slot on each side of the five consonants:
_ C _ C _ C _ C _ C _
You can choose which of those six slots to fill in $6 choose 3$ ways. And for each such choice there are $3!$ ways to put the three vowels.
Thus: $5! times (6 cdot 5 cdot 4)/(3 cdot 2 cdot 1) times 3! = 14400 $ ways.
answered Aug 23 at 1:24
David G. Stork
8,03121232
How do you figure 7 slots? There are 6.
– Paul Childs
Aug 23 at 1:27
You didn't do choosing properly. You should either get rid of the extra $3!$ or add a $(3 cdot 2 cdot 1)$ in the denominator.
– John Lou
Aug 23 at 2:41
@DavidG.Stork $5! times (6 cdot 5 cdot 4)/(3 cdot 2 cdot 1) times 3!=14400$ but not $5! times (6 cdot 5 cdot 4)/(3 cdot 2 cdot 1) times 3!ne86400$
– user572932
Aug 23 at 2:52
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Yes, your answer is correct.
There are $5$ consonants and $3$ vowels.
Vowels can be selected in $dbinom63$ ways.
The vowels be arranged in $3!$ ways
The consonants can be arranged in $5!$ ways.
In total we have $dbinom63times3!times5!=14400$ ways.
answered Aug 23 at 1:18
Key Flex
1
1
Thank you . I will accept your answer with green tick :)
– user122343
Aug 23 at 1:21
@user122343 You are welcome :)
– Key Flex
Aug 23 at 1:22
Sorry I thought I did. I just did it back.
– user122343
Aug 25 at 6:01
add a comment |Â
up vote
3
down vote
accepted
Yes, your answer is correct.
There are $5$ consonants and $3$ vowels.
Vowels can be selected in $dbinom63$ ways.
The vowels be arranged in $3!$ ways
The consonants can be arranged in $5!$ ways.
In total we have $dbinom63times3!times5!=14400$ ways.
answered Aug 23 at 1:18
Key Flex
1
1
Thank you . I will accept your answer with green tick :)
– user122343
Aug 23 at 1:21
@user122343 You are welcome :)
– Key Flex
Aug 23 at 1:22
Sorry I thought I did. I just did it back.
– user122343
Aug 25 at 6:01
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Yes, your answer is correct.
There are $5$ consonants and $3$ vowels.
Vowels can be selected in $dbinom63$ ways.
The vowels be arranged in $3!$ ways
The consonants can be arranged in $5!$ ways.
In total we have $dbinom63times3!times5!=14400$ ways.
answered Aug 23 at 1:18
Key Flex
1
Yes, your answer is correct.
There are $5$ consonants and $3$ vowels.
Vowels can be selected in $dbinom63$ ways.
The vowels be arranged in $3!$ ways
The consonants can be arranged in $5!$ ways.
In total we have $dbinom63times3!times5!=14400$ ways.
answered Aug 23 at 1:18
Key Flex
1
edited Aug 23 at 1:26
answered Aug 23 at 1:18
Key Flex
1
answered Aug 23 at 1:18
Key Flex
1
answered Aug 23 at 1:18
Key Flex
1
1
1
Thank you . I will accept your answer with green tick :)
– user122343
Aug 23 at 1:21
@user122343 You are welcome :)
– Key Flex
Aug 23 at 1:22
Sorry I thought I did. I just did it back.
– user122343
Aug 25 at 6:01
add a comment |Â
1
Thank you . I will accept your answer with green tick :)
– user122343
Aug 23 at 1:21
@user122343 You are welcome :)
– Key Flex
Aug 23 at 1:22
Sorry I thought I did. I just did it back.
– user122343
Aug 25 at 6:01
1
1
Thank you . I will accept your answer with green tick :)
– user122343
Aug 23 at 1:21
Thank you . I will accept your answer with green tick :)
– user122343
Aug 23 at 1:21
@user122343 You are welcome :)
– Key Flex
Aug 23 at 1:22
@user122343 You are welcome :)
– Key Flex
Aug 23 at 1:22
Sorry I thought I did. I just did it back.
– user122343
Aug 25 at 6:01
Sorry I thought I did. I just did it back.
– user122343
Aug 25 at 6:01
add a comment |Â
up vote
0
down vote
The consonants can be arranged in $5!$ ways. There are then six "slots" available for single vowels: one slot on each side of the five consonants:
_ C _ C _ C _ C _ C _
You can choose which of those six slots to fill in $6 choose 3$ ways. And for each such choice there are $3!$ ways to put the three vowels.
Thus: $5! times (6 cdot 5 cdot 4)/(3 cdot 2 cdot 1) times 3! = 14400 $ ways.
answered Aug 23 at 1:24
David G. Stork
8,03121232
How do you figure 7 slots? There are 6.
– Paul Childs
Aug 23 at 1:27
You didn't do choosing properly. You should either get rid of the extra $3!$ or add a $(3 cdot 2 cdot 1)$ in the denominator.
– John Lou
Aug 23 at 2:41
@DavidG.Stork $5! times (6 cdot 5 cdot 4)/(3 cdot 2 cdot 1) times 3!=14400$ but not $5! times (6 cdot 5 cdot 4)/(3 cdot 2 cdot 1) times 3!ne86400$
– user572932
Aug 23 at 2:52
add a comment |Â
up vote
0
down vote
The consonants can be arranged in $5!$ ways. There are then six "slots" available for single vowels: one slot on each side of the five consonants:
_ C _ C _ C _ C _ C _
You can choose which of those six slots to fill in $6 choose 3$ ways. And for each such choice there are $3!$ ways to put the three vowels.
Thus: $5! times (6 cdot 5 cdot 4)/(3 cdot 2 cdot 1) times 3! = 14400 $ ways.
answered Aug 23 at 1:24
David G. Stork
8,03121232
How do you figure 7 slots? There are 6.
– Paul Childs
Aug 23 at 1:27
You didn't do choosing properly. You should either get rid of the extra $3!$ or add a $(3 cdot 2 cdot 1)$ in the denominator.
– John Lou
Aug 23 at 2:41
@DavidG.Stork $5! times (6 cdot 5 cdot 4)/(3 cdot 2 cdot 1) times 3!=14400$ but not $5! times (6 cdot 5 cdot 4)/(3 cdot 2 cdot 1) times 3!ne86400$
– user572932
Aug 23 at 2:52
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The consonants can be arranged in $5!$ ways. There are then six "slots" available for single vowels: one slot on each side of the five consonants:
_ C _ C _ C _ C _ C _
You can choose which of those six slots to fill in $6 choose 3$ ways. And for each such choice there are $3!$ ways to put the three vowels.
Thus: $5! times (6 cdot 5 cdot 4)/(3 cdot 2 cdot 1) times 3! = 14400 $ ways.
answered Aug 23 at 1:24
David G. Stork
8,03121232
The consonants can be arranged in $5!$ ways. There are then six "slots" available for single vowels: one slot on each side of the five consonants:
_ C _ C _ C _ C _ C _
You can choose which of those six slots to fill in $6 choose 3$ ways. And for each such choice there are $3!$ ways to put the three vowels.
Thus: $5! times (6 cdot 5 cdot 4)/(3 cdot 2 cdot 1) times 3! = 14400 $ ways.
answered Aug 23 at 1:24
David G. Stork
8,03121232
edited Aug 23 at 3:47
answered Aug 23 at 1:24
David G. Stork
8,03121232
answered Aug 23 at 1:24
David G. Stork
8,03121232
answered Aug 23 at 1:24
David G. Stork
8,03121232
8,03121232
How do you figure 7 slots? There are 6.
– Paul Childs
Aug 23 at 1:27
You didn't do choosing properly. You should either get rid of the extra $3!$ or add a $(3 cdot 2 cdot 1)$ in the denominator.
– John Lou
Aug 23 at 2:41
@DavidG.Stork $5! times (6 cdot 5 cdot 4)/(3 cdot 2 cdot 1) times 3!=14400$ but not $5! times (6 cdot 5 cdot 4)/(3 cdot 2 cdot 1) times 3!ne86400$
– user572932
Aug 23 at 2:52
add a comment |Â
How do you figure 7 slots? There are 6.
– Paul Childs
Aug 23 at 1:27
You didn't do choosing properly. You should either get rid of the extra $3!$ or add a $(3 cdot 2 cdot 1)$ in the denominator.
– John Lou
Aug 23 at 2:41
@DavidG.Stork $5! times (6 cdot 5 cdot 4)/(3 cdot 2 cdot 1) times 3!=14400$ but not $5! times (6 cdot 5 cdot 4)/(3 cdot 2 cdot 1) times 3!ne86400$
– user572932
Aug 23 at 2:52
How do you figure 7 slots? There are 6.
– Paul Childs
Aug 23 at 1:27
How do you figure 7 slots? There are 6.
– Paul Childs
Aug 23 at 1:27
You didn't do choosing properly. You should either get rid of the extra $3!$ or add a $(3 cdot 2 cdot 1)$ in the denominator.
– John Lou
Aug 23 at 2:41
You didn't do choosing properly. You should either get rid of the extra $3!$ or add a $(3 cdot 2 cdot 1)$ in the denominator.
– John Lou
Aug 23 at 2:41
@DavidG.Stork $5! times (6 cdot 5 cdot 4)/(3 cdot 2 cdot 1) times 3!=14400$ but not $5! times (6 cdot 5 cdot 4)/(3 cdot 2 cdot 1) times 3!ne86400$
– user572932
Aug 23 at 2:52
@DavidG.Stork $5! times (6 cdot 5 cdot 4)/(3 cdot 2 cdot 1) times 3!=14400$ but not $5! times (6 cdot 5 cdot 4)/(3 cdot 2 cdot 1) times 3!ne86400$
– user572932
Aug 23 at 2:52
add a comment |Â
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