# How many arrangements of the word ACHIEVE are there if the letters A & E were to beWatch

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Thread starter 1 month ago
#1
How many arrangements of the word ACHIEVE are there if the letters A & E were to be together ?
0
1 month ago
#2
(Original post by Leah.J)
How many arrangements of the word ACHIEVE are there if the letters A & E were to be together ?
Any thoughts? You've done some similar questions, so what is different this time?
0
1 month ago
#3
(Original post by Leah.J)
How many arrangements of the word ACHIEVE are there if the letters A & E were to be together ?
You've already asked this question once - https://www.thestudentroom.co.uk/sho....php?t=6229728

Ignore this.
Last edited by ghostwalker; 1 month ago
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Thread starter 1 month ago
#4
(Original post by ghostwalker)
You've already asked this question once - https://www.thestudentroom.co.uk/sho....php?t=6229728
No but it in this case there 2 E letters
So I’m sure not whether 6!/2! *2! works
Shouldn’t the chance that an A and E be together be higher than the chance that an A and an I ( for example )are together ?
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1 month ago
#5
(Original post by Leah.J)
No but it in this case there 2 E letters
So I’m sure not whether 6!/2! *2! works
Shouldn’t the chance that an A and E be together be higher than the chance that an A and an I ( for example )are together ?
First forget about there are two Es. If one E was Z, how many combinations would there be with AE together? Treat "AE" as a single letter.
Last edited by mqb2766; 1 month ago
0
1 month ago
#6
(Original post by Leah.J)
No but it in this case there 2 E letters
So I’m sure not whether 6!/2! *2! works
Shouldn’t the chance that an A and E be together be higher than the chance that an A and an I ( for example )are together ?
Yes, my apologies, it is a different question.
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Thread starter 1 month ago
#7
(Original post by mqb2766)
First forget about there are two Es. If one E was Z, how many combinations would there be with AE together? Treat "AE" as a single letter.
In this case it would be 6! * 2!
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1 month ago
#8
(Original post by Leah.J)
In this case it would be 6! * 2!
Yes. Now replace the Z by E, how does this differ?
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Thread starter 1 month ago
#9
(Original post by mqb2766)
Yes. Now replace the Z by E, how does this differ?
I don't know, does it not ?
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1 month ago
#10
(Original post by Leah.J)
I don't know, does it not ?
How many ways could you have
AEECHIV
With the double E? How should the previous 6!2! be adjusted?
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Thread starter 1 month ago
#11
(Original post by mqb2766)
How many ways could you have
AEECHIV
With the double E? How should the previous 6!2! be adjusted?
I'm really not the best at this but to get rid of the E1E2 = E2E1 problem we should divide by 2, I know that much.
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1 month ago
#12
(Original post by Leah.J)
I'm really not the best at this but to get rid of the E1E2 = E2E1 problem we should divide by 2, I know that much.
How about working out a simple example
AEEV
Which has the same rules, an A and an E must appear together.
How many combinations are there and how many repeats? What pattern does those repeats have?
0
1 month ago
#13
(Original post by Leah.J)
...
Since there's been no movement for a day, I though this might help.

When you initially worked out the number of entries by lumping "A" and "E" together as one item and multiplying by two you were double counting some entries. If we now split it out into the two cases, "AE" and "EA", and put it on a Venn diagram, perhaps it may be clearer which ones are being double counted:

Last edited by ghostwalker; 1 month ago
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Thread starter 1 month ago
#14
(Original post by mqb2766)
How about working out a simple example
AEEV
Which has the same rules, an A and an E must appear together.
How many combinations are there and how many repeats? What pattern does those repeats have?
sorry I completely forgot I asked this question, by finding all the possible outcomes, I got 8, but I don't know how to solve this. I don't know why I'm finding it hard. There was another question that said : Find how many different arrangements there are of the nine letters in the words GOLD MEDAL if the two letters D come first and the two letters L come last.

and I got the right answer by doing 2*1*5*4*3*2*2*1 and then dividing by 2! and 2! again
Last edited by Leah.J; 1 month ago
0
Thread starter 1 month ago
#15
(Original post by Leah.J)
sorry I completely forgot I asked this question, by finding all the possible outcomesr, I got 8, but I don't know how to solve this. I don't know why I'm finding it hard. There was another question that said : Find how many different arrangements there are of the nine letters in the words GOLD MEDAL if the two letters D come first and the two letters L come last.

and I got the right answer by doing 2*1*5*4*3*2*2*1 and then dividing by 2! and 2! again
I tried doing this :
I put 7 empty spaces next to each other and I said there are 2 scenarios : 1. an A comes first and 2. an E comes first.
if an A comes first then it's 1*2*5*4*3*2*1
if an E comes first then its 2*1*5*4*3*2*1
what should I do next ? divide each by 2! and add ?
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Thread starter 1 month ago
#16
(Original post by ghostwalker)
Since there's been no movement for a day, I though this might help.

When you initially worked out the number of entries by lumping "A" and "E" together as one item and multiplying by two you were double counting some entries. If we now split it out into the two cases, "AE" and "EA", and put it on a Venn diagram, perhaps it may be clearer which ones are being double counted:

you mean by doing 6!*2!/2! ?
Can you walk me through the approach please.
0
1 month ago
#17
You only double count when you have the substring
EAE
So subttact the number of times that occurs.
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Thread starter 1 month ago
#18
(Original post by mqb2766)
You only double count when you have the substring
EAE
So subttact the number of times that occurs.
how does that take into account the 2 E's ? CHE1E2AVI is the same as CHE2E1AVI
0
1 month ago
#19
(Original post by Leah.J)
how does that take into account the 2 E's ? CHE1E2AVI is the same as CHE2E1AVI
AEEV
Is simpler to analyse. Pick one E for AE or EA and you have EV left.
So you can have only one of
VEEA
when you do the original count and ttreat EA a single letter. There is no double counting for this arrangement.
The only double counting occurs for things like
VEAE
Where the Es surround the A and could be generated from the free E or the AE/EA.

For this case you should get 12 in the original count (2*3!). Two of which are repeats (2!) so 10 in total..
VEAE
EAEV
AEEV
AEVE
VAEE
EVAE
EAVE
EEAV
EVEA
VEEA
Last edited by mqb2766; 1 month ago
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