From the letters in MAGOOSH, we are going to make three-letter "words.

Question

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

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

chetan2u · Accepted Answer

(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

merimanukyan · Answer

I am confused a bit. why we can't consider 7 letters from the beginning and arrange them? I mean 7*6*5 = 210?

chetan2u · Answer

If you take 7 letters it will consist of both O. 
Say the letters are M, A,G,O1,O2,S,H
So these 7*6*5 will consist of O1MA and O2MA. But both are same, OMA, so there are repetition in 7*6*5.

RohitSaluja · Answer

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

UzairSohail · Answer

Hello. Did you get any answer on this? Because I was thinking the same. Why can't we solve this question by 7P3/2!?

GmatPoint · Answer

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.

UzairSohail · Answer

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.

chetan2u · Answer

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

UzairSohail · Answer

Oh I see. Yeah yeah, I understand it now. Thank you so much for pointing my mistake! It's a big help

onlymalapink · Answer

why isnt there a 2! to repeated o equation they can have different words for example ooa so aoa

kayarat600 · Answer

Permutation is used for this expres purpose

HarshavardhanR · Answer

There are two possible use cases ->

(1) All letters different

From 6 possible letters, 3 are chosen, and then, these 3 are arranged -> 6C3 x 3! = 6 x 5 x 4 = 120

(2) Two Os, one other letter

O is already picked. 5C1 ways to pick the other letter and 3!/2! ways to arrange (O, O, the other letter) -> 5C1 x 3!/2! = 15

Total = 120 + 15 = 135. Choice A.

---
Harsha

From the letters in MAGOOSH, we are going to make three-letter "words.

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From the letters in MAGOOSH, we are going to make three-letter "words.

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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