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sanemask
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Question about rearranging sequence with fixed duplicates 19 Oct 2015, 00:12
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Hi,

how many different ways can this fixed sequence be rearranged?
and what is the formula for this?

For 2 B and 2 G: There are 6 ways it can be rearranged

BGBG
BBGG
GGBB
GBBG
BGGB
GBGB

Don't Tell me it's N!/(N-R)!R! because It just incidentally has the same result for this case.


To elaborate, I will introduce more several similar cases:

For 2 G and 1 B: There are 3 different ways

GGB
GBG
BGG

For 3 G and 1 B: There are 4 different ways

GGGB
BGGG
GBGG
GGBG

Which N!/(N-R)!R! cannot come to a result

PS. If i were ever mistaken, N!/(N-R)!R! is for entirely different case right? It's for combinations like lottery scenario that shows how many different way can slots of numbers be picking out while order doesn't matter



Or Furthermore, for 2 A, 2 B and 2 C...... who knows for this one, I don't even want to try lol

AABBCC
so on.....

Thank you,



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ENGRTOMBA2018
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Question about rearranging sequence with fixed duplicates 19 Oct 2015, 05:15
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sanemask
Hi,

how many different ways can this fixed sequence be rearranged?
and what is the formula for this?

For 2 B and 2 G: There are 6 ways it can be rearranged

BGBG
BBGG
GGBB
GBBG
BGGB
GBGB

Don't Tell me it's N!/(N-R)!R! because It just incidentally has the same result for this case.


To elaborate, I will introduce more several similar cases:

For 2 G and 1 B: There are 3 different ways

GGB
GBG
BGG

For 3 G and 1 B: There are 4 different ways

GGGB
BGGG
GBGG
GGBG

Which N!/(N-R)!R! cannot come to a result

PS. If i were ever mistaken, N!/(N-R)!R! is for entirely different case right? It's for combinations like lottery scenario that shows how many different way can slots of numbers be picking out while order doesn't matter



Or Furthermore, for 2 A, 2 B and 2 C...... who knows for this one, I don't even want to try lol

AABBCC
so on.....

Thank you,
Show more

What you are asking is "how at ARRANGE...". The standard formula for such a thing is as follows

Case 1: 2 B and 2G , number of arrangements = 4!/(2!*2!)=6 ---> 4! = number of arrangements of 4 elements but as 2 Bs and 2 Gs are exactly the same so you need to divide 4! by 2*2! to account for this "sameness" and to eliminate repeated arrangements.

Case 2: 3 G and 1 B, again the same principle = 4!/ (3!) = 4 arrangements

Case 3: 2A, 2B, 2C = 6!/(2!*2*2!) = 720/8 = 90 ways

Additionally, the formulae for arrangements and combinations is as follows:

For arrangement of AA...A (m times)BBBBB...(n times)CCCCC....(p times)DDDDD...(q times) etc , the number of arrangements = (m+n+p+q)!/(m!*n!*p!*q!)

For choosing p elements out of a total of n objects = nCp = n!/(p!*(n-p)!)

Hope this helps.
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sanemask
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Question about rearranging sequence with fixed duplicates 19 Oct 2015, 20:29
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Engr2012
sanemask
Hi,

how many different ways can this fixed sequence be rearranged?
and what is the formula for this?

For 2 B and 2 G: There are 6 ways it can be rearranged

BGBG
BBGG
GGBB
GBBG
BGGB
GBGB

Don't Tell me it's N!/(N-R)!R! because It just incidentally has the same result for this case.


To elaborate, I will introduce more several similar cases:

For 2 G and 1 B: There are 3 different ways

GGB
GBG
BGG

For 3 G and 1 B: There are 4 different ways

GGGB
BGGG
GBGG
GGBG

Which N!/(N-R)!R! cannot come to a result

PS. If i were ever mistaken, N!/(N-R)!R! is for entirely different case right? It's for combinations like lottery scenario that shows how many different way can slots of numbers be picking out while order doesn't matter



Or Furthermore, for 2 A, 2 B and 2 C...... who knows for this one, I don't even want to try lol

AABBCC
so on.....

Thank you,

What you are asking is "how at ARRANGE...". The standard formula for such a thing is as follows

Case 1: 2 B and 2G , number of arrangements = 4!/(2!*2!)=6 ---> 4! = number of arrangements of 4 elements but as 2 Bs and 2 Gs are exactly the same so you need to divide 4! by 2*2! to account for this "sameness" and to eliminate repeated arrangements.

Case 2: 3 G and 1 B, again the same principle = 4!/ (3!) = 4 arrangements

Case 3: 2A, 2B, 2C = 6!/(2!*2*2!) = 720/8 = 90 ways

Additionally, the formulae for arrangements and combinations is as follows:

For arrangement of AA...A (m times)BBBBB...(n times)CCCCC....(p times)DDDDD...(q times) etc , the number of arrangements = (m+n+p+q)!/(m!*n!*p!*q!)

For choosing p elements out of a total of n objects = nCp = n!/(p!*(n-p)!)

Hope this helps.
Show more



Thank you sir,

and one more thing, what is the formula called?

I know that the last one is called combination.
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ENGRTOMBA2018
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3,889
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Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE:Engineering (Aerospace and Defense)
Products:
My Reviews
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
Posts: 2,319
Kudos: 3,889
Question about rearranging sequence with fixed duplicates 20 Oct 2015, 03:41
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sanemask
Engr2012
sanemask
Hi,

how many different ways can this fixed sequence be rearranged?
and what is the formula for this?

For 2 B and 2 G: There are 6 ways it can be rearranged

BGBG
BBGG
GGBB
GBBG
BGGB
GBGB

Don't Tell me it's N!/(N-R)!R! because It just incidentally has the same result for this case.


To elaborate, I will introduce more several similar cases:

For 2 G and 1 B: There are 3 different ways

GGB
GBG
BGG

For 3 G and 1 B: There are 4 different ways

GGGB
BGGG
GBGG
GGBG

Which N!/(N-R)!R! cannot come to a result

PS. If i were ever mistaken, N!/(N-R)!R! is for entirely different case right? It's for combinations like lottery scenario that shows how many different way can slots of numbers be picking out while order doesn't matter



Or Furthermore, for 2 A, 2 B and 2 C...... who knows for this one, I don't even want to try lol

AABBCC
so on.....

Thank you,

What you are asking is "how at ARRANGE...". The standard formula for such a thing is as follows

Case 1: 2 B and 2G , number of arrangements = 4!/(2!*2!)=6 ---> 4! = number of arrangements of 4 elements but as 2 Bs and 2 Gs are exactly the same so you need to divide 4! by 2*2! to account for this "sameness" and to eliminate repeated arrangements.

Case 2: 3 G and 1 B, again the same principle = 4!/ (3!) = 4 arrangements

Case 3: 2A, 2B, 2C = 6!/(2!*2*2!) = 720/8 = 90 ways

Additionally, the formulae for arrangements and combinations is as follows:

For arrangement of AA...A (m times)BBBBB...(n times)CCCCC....(p times)DDDDD...(q times) etc , the number of arrangements = (m+n+p+q)!/(m!*n!*p!*q!)

For choosing p elements out of a total of n objects = nCp = n!/(p!*(n-p)!)

Hope this helps.



Thank you sir,

and one more thing, what is the formula called?

I know that the last one is called combination.
Show more


When you talk about arrangements, it is called permutations and when you talk about selections it is known as combinations.

Hope this helps.
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Question about rearranging sequence with fixed duplicates 20 Oct 2015, 09:09
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sanemask
Engr2012
sanemask
Hi,

how many different ways can this fixed sequence be rearranged?
and what is the formula for this?

For 2 B and 2 G: There are 6 ways it can be rearranged

BGBG
BBGG
GGBB
GBBG
BGGB
GBGB

Don't Tell me it's N!/(N-R)!R! because It just incidentally has the same result for this case.


To elaborate, I will introduce more several similar cases:

For 2 G and 1 B: There are 3 different ways

GGB
GBG
BGG

For 3 G and 1 B: There are 4 different ways

GGGB
BGGG
GBGG
GGBG

Which N!/(N-R)!R! cannot come to a result

PS. If i were ever mistaken, N!/(N-R)!R! is for entirely different case right? It's for combinations like lottery scenario that shows how many different way can slots of numbers be picking out while order doesn't matter



Or Furthermore, for 2 A, 2 B and 2 C...... who knows for this one, I don't even want to try lol

AABBCC
so on.....

Thank you,

What you are asking is "how at ARRANGE...". The standard formula for such a thing is as follows

Case 1: 2 B and 2G , number of arrangements = 4!/(2!*2!)=6 ---> 4! = number of arrangements of 4 elements but as 2 Bs and 2 Gs are exactly the same so you need to divide 4! by 2*2! to account for this "sameness" and to eliminate repeated arrangements.

Case 2: 3 G and 1 B, again the same principle = 4!/ (3!) = 4 arrangements

Case 3: 2A, 2B, 2C = 6!/(2!*2*2!) = 720/8 = 90 ways

Additionally, the formulae for arrangements and combinations is as follows:

For arrangement of AA...A (m times)BBBBB...(n times)CCCCC....(p times)DDDDD...(q times) etc , the number of arrangements = (m+n+p+q)!/(m!*n!*p!*q!)

For choosing p elements out of a total of n objects = nCp = n!/(p!*(n-p)!)

Hope this helps.



Thank you sir,

and one more thing, what is the formula called?

I know that the last one is called combination.
Show more

sanemask -

This formula is for Permutations with repeating elements. You will use this formula for questions whenever there are multiple instances of one or more of the items to be arranged (such as, "how many ways can the letters of the MISSISSIPPI be arranged", or what is the probability of getting exactly 3 Heads when flipping a coin 5 times).



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!