Last visit was: 06 May 2024, 19:21 It is currently 06 May 2024, 19:21

Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
SORT BY:
Kudos
Tags:
Show Tags
Hide Tags
Manager
Manager
Joined: 16 Dec 2018
Posts: 60
Own Kudos [?]: 101 [24]
Given Kudos: 48
Send PM
Most Helpful Reply
RC & DI Moderator
Joined: 02 Aug 2009
Status:Math and DI Expert
Posts: 11216
Own Kudos [?]: 32316 [6]
Given Kudos: 301
Send PM
General Discussion
RC & DI Moderator
Joined: 02 Aug 2009
Status:Math and DI Expert
Posts: 11216
Own Kudos [?]: 32316 [2]
Given Kudos: 301
Send PM
Manager
Manager
Joined: 02 Aug 2020
Posts: 216
Own Kudos [?]: 85 [2]
Given Kudos: 254
Location: India
Concentration: General Management, Healthcare
Schools: HEC'22 (J)
GMAT 1: 720 Q49 V40
GPA: 3.8
WE:Consulting (Health Care)
Send PM
Re: From the letters in MAGOOSH, we are going to make three-letter "words. [#permalink]
2
Kudos
chetan2u wrote:
hudacse6 wrote:
From the letters in MAGOOSH, we are going to make three-letter "words." Any set of three letters counts as a word, and different arrangements of the same three letters (such as "MAG" and "AGM") count as different words. How many different three-letter words can be made from the seven letters in MAGOOSH?

A. 135
B. 170
C. 123
D. 121
E. 720


(I) Take all different letters.. M, A, G, O, S and H, so 6 letters...
They can be arranged in 6*5*4=120 ways
(II) two are O and third any of remaining 5..
So \(5*\frac{3!}{2!} =15\) as each combination can be arranged in \(\frac{ 3!}{2!}\) ways

Total 120+15=135

A



Why cant we write it as 7P3/2!, dividing by 2! for two O's
Senior Manager
Senior Manager
Joined: 02 Jan 2022
Posts: 263
Own Kudos [?]: 95 [2]
Given Kudos: 3
GMAT 1: 760 Q50 V42
Send PM
Re: From the letters in MAGOOSH, we are going to make three-letter "words. [#permalink]
2
Bookmarks
There are 6 different letters with repetition of one letter, i.e. O.

Let the 3-letter words have all the letters different.

When all the letters are different, the 3 letters can be selected in 6C3 ways. Since the letters can arrange among themselves in 3! ways,

The number of ways to select 3 letters out of 6 = 6C3*3! = 120

When one letter is repeated, two out of the three letters will be O. The last letter can be selected in 5 ways and the letters can arrange themselves in 3!/2! = 3 ways.

Thus, the number of 3-letter words with the repetition of O = 5*3 = 15

Thus, total letters = 120+15 = 135

Thus, the correct option is A.
Intern
Intern
Joined: 13 Jun 2019
Posts: 4
Own Kudos [?]: 1 [1]
Given Kudos: 6
Send PM
From the letters in MAGOOSH, we are going to make three-letter "words. [#permalink]
1
Kudos
chetan2u wrote:
hudacse6 wrote:
From the letters in MAGOOSH, we are going to make three-letter "words." Any set of three letters counts as a word, and different arrangements of the same three letters (such as "MAG" and "AGM") count as different words. How many different three-letter words can be made from the seven letters in MAGOOSH?

A. 135
B. 170
C. 123
D. 121
E. 720


(I) Take all different letters.. M, A, G, O, S and H, so 6 letters...
They can be arranged in 6*5*3=120 ways
(II) two are O and third any of remaining 5..
So \(5*\frac{3!}{2!} =15\) as each combination can be arranged in \(\frac{ 3!}{2!}\) ways

Total 120+15=135

A

I am confused a bit. why we can't consider 7 letters from the beginning and arrange them? I mean 7*6*5 = 210?
Intern
Intern
Joined: 13 Jun 2019
Posts: 4
Own Kudos [?]: 1 [0]
Given Kudos: 6
Send PM
Re: From the letters in MAGOOSH, we are going to make three-letter "words. [#permalink]
Thank you! That makes sense!

Posted from my mobile device
Intern
Intern
Joined: 31 Jul 2021
Posts: 18
Own Kudos [?]: 15 [0]
Given Kudos: 62
Location: Pakistan
GMAT 1: 730 Q48 V42
GPA: 3.31
Send PM
Re: From the letters in MAGOOSH, we are going to make three-letter "words. [#permalink]
RohitSaluja wrote:
chetan2u wrote:
hudacse6 wrote:
From the letters in MAGOOSH, we are going to make three-letter "words." Any set of three letters counts as a word, and different arrangements of the same three letters (such as "MAG" and "AGM") count as different words. How many different three-letter words can be made from the seven letters in MAGOOSH?

A. 135
B. 170
C. 123
D. 121
E. 720


(I) Take all different letters.. M, A, G, O, S and H, so 6 letters...
They can be arranged in 6*5*4=120 ways
(II) two are O and third any of remaining 5..
So \(5*\frac{3!}{2!} =15\) as each combination can be arranged in \(\frac{ 3!}{2!}\) ways

Total 120+15=135

A



Why cant we write it as 7P3/2!, dividing by 2! for two O's


Hello. Did you get any answer on this? Because I was thinking the same. Why can't we solve this question by 7P3/2!?
Intern
Intern
Joined: 31 Jul 2021
Posts: 18
Own Kudos [?]: 15 [0]
Given Kudos: 62
Location: Pakistan
GMAT 1: 730 Q48 V42
GPA: 3.31
Send PM
Re: From the letters in MAGOOSH, we are going to make three-letter "words. [#permalink]
That's alright, I understand this solution. But my question is, why can't this be solved by 7P3/2!? What is the flaw in this strategy? Since we are choosing 3 letters from 7 letters, using permutation because order matters, and dividing the result by 2! because one letter repeats twice.
RC & DI Moderator
Joined: 02 Aug 2009
Status:Math and DI Expert
Posts: 11216
Own Kudos [?]: 32316 [0]
Given Kudos: 301
Send PM
Re: From the letters in MAGOOSH, we are going to make three-letter "words. [#permalink]
Expert Reply
UzairSohail wrote:
That's alright, I understand this solution. But my question is, why can't this be solved by 7P3/2!? What is the flaw in this strategy? Since we are choosing 3 letters from 7 letters, using permutation because order matters, and dividing the result by 2! because one letter repeats twice.



You divide by 2! when all the solutions have repetitions but is it so here in all cases. NO

If the letter is MAG, then there is no repetition. So you are subtracting more cases than you should by dividing everything by 2!.

What if we want to move fwd on 7*6*5 or 210 cases.

There are two types of repetition.

1) When the word contains exactly one O.
The other two can be chosen in 5C2 or 10 ways. But each way can be arranged in 3! ways.
Thus, total 10*3! Or 60 ways.

2) When the word contains 2 Os.
The other one can be chosen in 5C1 or 5 ways. But each way can be arranged in 3!/2! ways.
Thus, total 5*3!/2! Or 15 ways.

Ways that are repeated 60+15 or 75 ways.


Total = 210-75 = 135 ways

Hope it helps
Intern
Intern
Joined: 31 Jul 2021
Posts: 18
Own Kudos [?]: 15 [0]
Given Kudos: 62
Location: Pakistan
GMAT 1: 730 Q48 V42
GPA: 3.31
Send PM
Re: From the letters in MAGOOSH, we are going to make three-letter "words. [#permalink]
chetan2u wrote:
UzairSohail wrote:
That's alright, I understand this solution. But my question is, why can't this be solved by 7P3/2!? What is the flaw in this strategy? Since we are choosing 3 letters from 7 letters, using permutation because order matters, and dividing the result by 2! because one letter repeats twice.



You divide by 2! when all the solutions have repetitions but is it so here in all cases. NO

If the letter is MAG, then there is no repetition. So you are subtracting more cases than you should by dividing everything by 2!.

What if we want to move fwd on 7*6*5 or 210 cases.

There are two types of repetition.

1) When the word contains exactly one O.
The other two can be chosen in 5C2 or 10 ways. But each way can be arranged in 3! ways.
Thus, total 10*3! Or 60 ways.

2) When the word contains 2 Os.
The other one can be chosen in 5C1 or 5 ways. But each way can be arranged in 3!/2! ways.
Thus, total 5*3!/2! Or 15 ways.

Ways that are repeated 60+15 or 75 ways.


Total = 210-75 = 135 ways

Hope it helps


Oh I see. Yeah yeah, I understand it now. Thank you so much for pointing my mistake! It's a big help :)
User avatar
Non-Human User
Joined: 09 Sep 2013
Posts: 32796
Own Kudos [?]: 827 [0]
Given Kudos: 0
Send PM
Re: From the letters in MAGOOSH, we are going to make three-letter "words. [#permalink]
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
GMAT Club Bot
Re: From the letters in MAGOOSH, we are going to make three-letter "words. [#permalink]
Moderators:
Math Expert
93060 posts
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne