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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?
4meonly wrote:Part A would be 8!/3!*2!b.kurland wrote:for part A, the logic seems right but 8!/3!2! = 3360
quote]
Agree
How to solve last 2 problems?
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...
sudhir3127 wrote: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.4meonly wrote:Part A would be 8!/3!*2!b.kurland wrote:for part A, the logic seems right but 8!/3!2! = 3360
quote]
Agree
How to solve last 2 problems?
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...
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?
sudhir3127 wrote:Sudhir,4meonly wrote:Part A would be 8!/3!*2!b.kurland wrote:for part A, the logic seems right but 8!/3!2! = 3360
quote]
Agree
How to solve last 2 problems?
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...
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).
sudhir3127 wrote: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)4meonly wrote:Part A would be 8!/3!*2!b.kurland wrote:for part A, the logic seems right but 8!/3!2! = 3360
quote]
Agree
How to solve last 2 problems?
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...
Stockmoose16 wrote: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.sudhir3127 wrote:Sudhir,4meonly wrote:Part A would be 8!/3!*2!b.kurland wrote:for part A, the logic seems right but 8!/3!2! = 3360
quote]
Agree
How to solve last 2 problems?
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...
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).
Stockmoose16 wrote: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.sudhir3127 wrote: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)4meonly wrote:Part A would be 8!/3!*2!b.kurland wrote:for part A, the logic seems right but 8!/3!2! = 3360
quote]
Agree
How to solve last 2 problems?
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...
pseudononymous 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:Stockmoose16 wrote: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.sudhir3127 wrote: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)4meonly wrote:Part A would be 8!/3!*2!b.kurland wrote:for part A, the logic seems right but 8!/3!2! = 3360
quote]
Agree
How to solve last 2 problems?
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...
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! = 1200Stockmoose16 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?
Why would you add back the instances with 3 consecutive U's?pseudononymous wrote:No, I think it's 8!/(3!*2!) - 7!/2! + 6!/2! = 1200Stockmoose16 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?
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