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805+ (Hard)|   Combinations|      
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Bunuel
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A “phrase” is made up of the first seven letters of the alphabet so 22 Feb 2022, 05:30
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43% (02:30) correct 57% (02:30) wrong based on 150 sessions

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A “phrase” is made up of the first seven letters of the alphabet so that each letter appears exactly once. The “phrase” must contain at least two “words”, which must contain at least two letters. For instance, ABC DE FG and AB CDE FG are distinct “phrases”. How many distinct “phrases” result given the conditions above?

(A) 5040

(B) 10,080

(C) 7!(5)

(D) 7!(6)

(E) 7!(7)


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A “phrase” is made up of the first seven letters of the alphabet so 24 Feb 2022, 04:06
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Bunuel
A “phrase” is made up of the first seven letters of the alphabet so that each letter appears exactly once. The “phrase” must contain at least two “words”, which must contain at least two letters. For instance, ABC DE FG and AB CDE FG are distinct “phrases”. How many distinct “phrases” result given the conditions above?

(A) 5040

(B) 10,080

(C) 7!(5)

(D) 7!(6)

(E) 7!(7)
Show more


TWO Groups: Let us give one each to the two groups x and y, then x+y=5 will give (n-1)C(r-1) = (5-1)C(2-1) = 4C1 = 4
THREE Groups: Let us give one each to the three groups x, y and z, then x+y+z=4 will give (4-1)C(3-1) = 3C2 = 3

Each of theses 7 ways of grouping have 7 alphabets that can be arranged in 7! ways => 7*7!

E
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A “phrase” is made up of the first seven letters of the alphabet so 24 Feb 2022, 23:44
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Bunuel
A “phrase” is made up of the first seven letters of the alphabet so that each letter appears exactly once. The “phrase” must contain at least two “words”, which must contain at least two letters. For instance, ABC DE FG and AB CDE FG are distinct “phrases”. How many distinct “phrases” result given the conditions above?



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THe first 2 options are 7!*1 and 7!*2 moving on,

THe combination of the letter can be either (1-3 group + 2-2 group ) 0r all (3-2 group and 1 member alone )

all of them has to be arranged in 7! ways

and the remaining combination can be either x+y= 5(3 group +2 group) = 4 (theory of partitioning ) + x+y+z= 3

Therefore IMO E
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A “phrase” is made up of the first seven letters of the alphabet so 01 Mar 2022, 09:09
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chetan2u
Bunuel
A “phrase” is made up of the first seven letters of the alphabet so that each letter appears exactly once. The “phrase” must contain at least two “words”, which must contain at least two letters. For instance, ABC DE FG and AB CDE FG are distinct “phrases”. How many distinct “phrases” result given the conditions above?

(A) 5040

(B) 10,080

(C) 7!(5)

(D) 7!(6)

(E) 7!(7)


TWO Groups: Let us give one each to the two groups x and y, then x+y=5 will give (n-1)C(r-1) = (5-1)C(2-1) = 4C1 = 4
THREE Groups: Let us give one each to the three groups x, y and z, then x+y+z=4 will give (4-1)C(3-1) = 3C2 = 3

Each of theses 7 ways of grouping have 7 alphabets that can be arranged in 7! ways => 7*7!

E
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Hi chetan2u

Some discrepancy I sense.

For TWO Groups : x+y=5 will give (n-1)C(r-1) = (5-1)C(2-1) = 4C1 = 4
Splitting the 4 combination, the distribution could be as follow : (2,5), (3,4), (4,3), (5,2)

The question is that x and y are not two distinct groups. So in that case pairs [(2,5) , (5,2)], and [(3,4), (4,3)] are equivalent.

Am I missing something ?
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A “phrase” is made up of the first seven letters of the alphabet so 02 Mar 2022, 09:02
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abhishekmayank
chetan2u
Bunuel
A “phrase” is made up of the first seven letters of the alphabet so that each letter appears exactly once. The “phrase” must contain at least two “words”, which must contain at least two letters. For instance, ABC DE FG and AB CDE FG are distinct “phrases”. How many distinct “phrases” result given the conditions above?

(A) 5040

(B) 10,080

(C) 7!(5)

(D) 7!(6)

(E) 7!(7)


TWO Groups: Let us give one each to the two groups x and y, then x+y=5 will give (n-1)C(r-1) = (5-1)C(2-1) = 4C1 = 4
THREE Groups: Let us give one each to the three groups x, y and z, then x+y+z=4 will give (4-1)C(3-1) = 3C2 = 3

Each of theses 7 ways of grouping have 7 alphabets that can be arranged in 7! ways => 7*7!

E

Hi chetan2u

Some discrepancy I sense.

For TWO Groups : x+y=5 will give (n-1)C(r-1) = (5-1)C(2-1) = 4C1 = 4
Splitting the 4 combination, the distribution could be as follow : (2,5), (3,4), (4,3), (5,2)

The question is that x and y are not two distinct groups. So in that case pairs [(2,5) , (5,2)], and [(3,4), (4,3)] are equivalent.

Am I missing something ?
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I stand to be corrected. Indeed these pairs, such as [(2,5) , (5,2)], are not equal as they will create different phrases..
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A “phrase” is made up of the first seven letters of the alphabet so 18 Sep 2022, 09:38
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7 alphabets are anyway going to be present which contributes to 7!

The other question is how many possible ways can we group the 7 alphabets. There are 7 possible ways.
(4,3), (3,4), (2,5), (5,2), (2,2,3), (2,3,2), (3,2,2).
The 7 alphabets are grouped according to one of the 7 possible ways mentioned above. Therefore, 7*7!

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A “phrase” is made up of the first seven letters of the alphabet so 22 Sep 2022, 11:10
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This may look harder than it seems to be. The below approach can be used:
To create a random phrase with 7 letters , we have a restriction of at-least 2 words per phrase with each word having at-least 2 letters.
So we are left with :
A,B,C,D,E,F,G = 3 word phrase or 2 word phrase

For 3 word phrases
Below are the possible combinations
2,2,3
2,3,2
3,2,2.
Thus a total of 3 ways. And each phrase can be arranged in 7! ways. Thus total no of 3 word phrases are 3*7!

For 2 word phrases
Below are the possible combinations
3,4 or 4,3
5,2 or 2,5.
Thus a total of 4 ways. And each phrase can be arranged in 7! ways. Thus total no of 2 word phrases are 4*7!.

Adding total no of 3 word and 2 word phrases gives 3*7! +4*7! = 7*7!.
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A “phrase” is made up of the first seven letters of the alphabet so 05 Sep 2025, 01:07
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A “phrase” is made up of the first seven letters of the alphabet so that each letter appears exactly once. The “phrase” must contain at least two “words”, which must contain at least two letters. For instance, ABC DE FG and AB CDE FG are distinct “phrases”.

How many distinct “phrases” result given the conditions above?

Case 1: 3L, 2L, 2L phrase
Number of ways to form the phrase = 3C1*7! = 3*7!

Case 2: 3L, 4L phrase
Number of ways to form the phrase = 2C1*7! = 2*7!

Case 3: 5L, 2L phrase
Number of ways to form the phrase = 2C1*7! = 2*7!

Total number of ways to form the phrase = 3*7! + 2*7! + 2*7! = 7*7!

IMO E
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