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Tough combinatrics problem

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Tough combinatrics problem

by Stockmoose16 » Fri Sep 12, 2008 6:20 pm
a. In how many ways can the letters in NUUNSUAL be arranged?

b. For the arrangements in part (a), how many have all three U's
together?

c. How many of the arrangements in part (a) have no consecutive U's?


Does anyone know how to figure out C?
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Re: Tough combinatrics problem

by amitdgr » Sat Sep 13, 2008 5:35 am
Stockmoose16 wrote: a. In how many ways can the letters in NUUNSUAL be arranged?
b. For the arrangements in part (a), how many have all three U's
together?
c. How many of the arrangements in part (a) have no consecutive U's?

Does anyone know how to figure out C?

A)Total number of distinct words possible-

NUUNSUAL is a 8 letter word. Letter N appears twice and U appears thrice
so it is 8! / (2! * 3!) = 3360 different words

B) all U's together - To find the arrangements where 3 U's appear together, we have to consider them to be a single unit.
For ease of understanding and to avoid mistakes, let us replace 3 U's by another letter say X
UUUNNSAL = XNNSAL

Now XNNSAL has 6 units (or letters) with 2 N's
so no of words = 6!/2! = 360 different words with 3 U's together

C) Words with no consecutive U's -

To find words with no consecutive U's = Number of different words that can be formed - Number of words with consecutive U's
words with no consecutive U's = 3360- 360 = 3000 words.

HTH

P.S. I EDITED THE SILLY CALCULATION ERROR I DID IN A AND THUS IN C
Last edited by amitdgr on Wed Sep 17, 2008 5:59 am, edited 1 time in total.
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by b.kurland » Sat Sep 13, 2008 11:39 am
for part A, the logic seems right but 8!/3!2! = 3360

and for part C, 360 only accounts for the instances of 3 consecutive U's and ignores the number of combinations in which there are 2 consecutive U's

i don't know how to solve for part C either... anyone?
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by 4meonly » Tue Sep 16, 2008 8:50 am
b.kurland wrote:for part A, the logic seems right but 8!/3!2! = 3360
/quote]

Agree

How to solve last 2 problems?
Last edited by 4meonly on Tue Sep 16, 2008 9:28 am, edited 1 time in total.
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by sudhir3127 » Tue Sep 16, 2008 9:19 am
4meonly wrote:
b.kurland wrote:for part A, the logic seems right but 8!/3!2! = 3360
quote]

Agree

How to solve last 2 problems?
Part A would be 8!/3!*2!

part B

all U's together - In all such Questions where a single unit is put together ... we need to consider that unit as one single entity....
thus it will be

NUUNSUAL: All U's willbe considered as one

hence we will have N N S A L and One Unit of "U"

thus its 6 alphabets...

the arrangement will be will

6!/ 2! ( 2! for 2 N's)

Part C.

Total words with no consecutive U's be

Arragements will all the words - Arrangements where all U's are together ..

which is nothing but Part A - Part B

8!/3!*2! - 6!/2!

3360- 360 = 3000

Hope this helps.. do let me know if u have any doubts...
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by b.kurland » Tue Sep 16, 2008 10:16 am
sudhir3127 wrote:
4meonly wrote:
b.kurland wrote:for part A, the logic seems right but 8!/3!2! = 3360
quote]

Agree

How to solve last 2 problems?
Part A would be 8!/3!*2!

part B

all U's together - In all such Questions where a single unit is put together ... we need to consider that unit as one single entity....
thus it will be

NUUNSUAL: All U's willbe considered as one

hence we will have N N S A L and One Unit of "U"

thus its 6 alphabets...

the arrangement will be will

6!/ 2! ( 2! for 2 N's)

Part C.

Total words with no consecutive U's be

Arragements will all the words - Arrangements where all U's are together ..

which is nothing but Part A - Part B

8!/3!*2! - 6!/2!

3360- 360 = 3000

Hope this helps.. do let me know if u have any doubts...
The problem is that Part B only groups all 3 U's together. This does = 6!/2! arrangements, or 360. However, this fails to account for the arrangements in which there are 2 consecutive U's.

So, the answer ought to be 3360 - (360 + # of arrangements with 2 consecutive U's). The # of 2 U arrangements is the tough part.

Any ideas how to solve?
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by pseudononymous » Tue Sep 16, 2008 10:33 am
I can do part (c) manually, but don't know how to do it with a formula:

U?U?U???
U?U??U??
U?U???U?
U?U????U
U??U?U??
U??U??U?
U??U???U
U???U?U?
U???U??U
U????U?U
?U?U?U??
?U?U??U?
?U?U???U
?U??U?U?
?U??U??U
?U???U?U
??U?U?U?
??U?U??U
??U??U?U
???U?U?U

20 configurations. The ?s, representing ALNNS, can be aranged 5!/2! = 60 times. 20 * 60 = 1200.

Anybody have an easier way to do this?
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by Stockmoose16 » Tue Sep 16, 2008 10:59 am
sudhir3127 wrote:
4meonly wrote:
b.kurland wrote:for part A, the logic seems right but 8!/3!2! = 3360
quote]

Agree

How to solve last 2 problems?
Part A would be 8!/3!*2!

part B

all U's together - In all such Questions where a single unit is put together ... we need to consider that unit as one single entity....
thus it will be

NUUNSUAL: All U's willbe considered as one

hence we will have N N S A L and One Unit of "U"

thus its 6 alphabets...

the arrangement will be will

6!/ 2! ( 2! for 2 N's)

Part C.

Total words with no consecutive U's be

Arragements will all the words - Arrangements where all U's are together ..

which is nothing but Part A - Part B

8!/3!*2! - 6!/2!

3360- 360 = 3000

Hope this helps.. do let me know if u have any doubts...
Sudhir,

I think your answer to part B is correct, BUT your logic is wrong. If you lock the U's together as one unit it would be UUUXXNXN. Or 5!/2!. Then you have to multiply by 6, for the number of arrangements that can be made of the Us. So the answer is 6!/2! (same answer, but you didn't account for the rearrangement).
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by Stockmoose16 » Tue Sep 16, 2008 11:05 am
sudhir3127 wrote:
4meonly wrote:
b.kurland wrote:for part A, the logic seems right but 8!/3!2! = 3360
quote]

Agree

How to solve last 2 problems?
Part A would be 8!/3!*2!

part B

all U's together - In all such Questions where a single unit is put together ... we need to consider that unit as one single entity....
thus it will be

NUUNSUAL: All U's willbe considered as one

hence we will have N N S A L and One Unit of "U"

thus its 6 alphabets...

the arrangement will be will

6!/ 2! ( 2! for 2 N's)

Part C.

Total words with no consecutive U's be

Arragements will all the words - Arrangements where all U's are together ..

which is nothing but Part A - Part B

8!/3!*2! - 6!/2!

3360- 360 = 3000

Hope this helps.. do let me know if u have any doubts...
The explanation for Part C is incorrect. In addition to subtracting out the instances with three consecutive U's, you need to subtract out the instances with 2 consecutive U's. The answer should be 480, NOT 3000. (3360-(6!/2!)-(7!/2!)=480)
Last edited by Stockmoose16 on Tue Sep 16, 2008 11:08 am, edited 1 time in total.
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by pseudononymous » Tue Sep 16, 2008 11:06 am
Stockmoose16 wrote:
sudhir3127 wrote:
4meonly wrote:
b.kurland wrote:for part A, the logic seems right but 8!/3!2! = 3360
quote]

Agree

How to solve last 2 problems?
Part A would be 8!/3!*2!

part B

all U's together - In all such Questions where a single unit is put together ... we need to consider that unit as one single entity....
thus it will be

NUUNSUAL: All U's willbe considered as one

hence we will have N N S A L and One Unit of "U"

thus its 6 alphabets...

the arrangement will be will

6!/ 2! ( 2! for 2 N's)

Part C.

Total words with no consecutive U's be

Arragements will all the words - Arrangements where all U's are together ..

which is nothing but Part A - Part B

8!/3!*2! - 6!/2!

3360- 360 = 3000

Hope this helps.. do let me know if u have any doubts...
Sudhir,

I think your answer to part B is correct, BUT your logic is wrong. If you lock the U's together as one unit it would be UUUXXNXN. Or 5!/2!. Then you have to multiply by 6, for the number of arrangements that can be made of the Us. So the answer is 6!/2! (same answer, but you didn't account for the rearrangement).
I think he's treating UUU as a single letter. It's the same thing as what you did, but just phrased differently. Both of you are correct.
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by pseudononymous » Tue Sep 16, 2008 11:18 am
Stockmoose16 wrote:
sudhir3127 wrote:
4meonly wrote:
b.kurland wrote:for part A, the logic seems right but 8!/3!2! = 3360
quote]

Agree

How to solve last 2 problems?
Part A would be 8!/3!*2!

part B

all U's together - In all such Questions where a single unit is put together ... we need to consider that unit as one single entity....
thus it will be

NUUNSUAL: All U's willbe considered as one

hence we will have N N S A L and One Unit of "U"

thus its 6 alphabets...

the arrangement will be will

6!/ 2! ( 2! for 2 N's)

Part C.

Total words with no consecutive U's be

Arragements will all the words - Arrangements where all U's are together ..

which is nothing but Part A - Part B

8!/3!*2! - 6!/2!

3360- 360 = 3000

Hope this helps.. do let me know if u have any doubts...
The explanation for Part C is incorrect. In addition to subtracting out the instances with three consecutive U's, you need to subtract out the instances with 2 consecutive U's. The answer should be 480, NOT 3000. (3360-(6!/2!)-(7!/2!)=480)
7!/2! doesn't give you the correct # of UU arrangements. It gives you the # of ways 7 letters can be arranged when 2 of the letters are the same. Also, if subtract both UUU and UU combos, you have to be sure not to subtract some arrangements twice.
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by Stockmoose16 » Tue Sep 16, 2008 11:27 am
pseudononymous wrote:
Stockmoose16 wrote:
sudhir3127 wrote:
4meonly wrote:
b.kurland wrote:for part A, the logic seems right but 8!/3!2! = 3360
quote]

Agree

How to solve last 2 problems?
Part A would be 8!/3!*2!

part B

all U's together - In all such Questions where a single unit is put together ... we need to consider that unit as one single entity....
thus it will be

NUUNSUAL: All U's willbe considered as one

hence we will have N N S A L and One Unit of "U"

thus its 6 alphabets...

the arrangement will be will

6!/ 2! ( 2! for 2 N's)

Part C.

Total words with no consecutive U's be

Arragements will all the words - Arrangements where all U's are together ..

which is nothing but Part A - Part B

8!/3!*2! - 6!/2!

3360- 360 = 3000

Hope this helps.. do let me know if u have any doubts...
The explanation for Part C is incorrect. In addition to subtracting out the instances with three consecutive U's, you need to subtract out the instances with 2 consecutive U's. The answer should be 480, NOT 3000. (3360-(6!/2!)-(7!/2!)=480)
7!/2! doesn't give you the correct # of UU arrangements. It gives you the # of ways 7 letters can be arranged when 2 of the letters are the same. Also, if subtract both UUU and UU combos, you have to be sure not to subtract some arrangements twice.
If you only subtract the places where the 3 Us are together, you're failing to count the places where 2 Us are together. Also, 7!/2! is derived from:

UUXXNXNU = 7!/2! ... you ADD this to the number of 3 U combinations to get the answer.

Can a GMAT INSTRUCTOR confirm this?
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by pseudononymous » Tue Sep 16, 2008 12:00 pm
Stockmoose16 wrote:
If you only subtract the places where the 3 Us are together, you're failing to count the places where 2 Us are together. Also, 7!/2! is derived from:

UUXXNXNU = 7!/2! ... you ADD this to the number of 3 U combinations to get the answer.

Can a GMAT INSTRUCTOR confirm this?
No, I think it's 8!/(3!*2!) - 7!/2! + 6!/2! = 1200
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by Stockmoose16 » Tue Sep 16, 2008 12:06 pm
pseudononymous wrote:
Stockmoose16 wrote:
If you only subtract the places where the 3 Us are together, you're failing to count the places where 2 Us are together. Also, 7!/2! is derived from:

UUXXNXNU = 7!/2! ... you ADD this to the number of 3 U combinations to get the answer.

Can a GMAT INSTRUCTOR confirm this?
No, I think it's 8!/(3!*2!) - 7!/2! + 6!/2! = 1200
Why would you add back the instances with 3 consecutive U's?
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by gmat579 » Tue Sep 16, 2008 12:58 pm
I believe that answer to part (c) is 8!/(3!*2!) - 7!/2!

Because, 7!/2! will give combinations of 3U's as well as 2U's.

Consider this,
Let, letter X =UU
Cases such as,
NXUNSAL, XUNNSAL etc. are infact 3U's together.

So subtracting 7!/2! must give all the combinations with no consecutive U's.
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