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permute with perfection

by sanju09 » Fri Feb 20, 2009 5:24 am
A 4-letter code word consists of letters A, B, and C. If the code includes all the three letters, how many such codes are possible?

(A) 72
(B) 48
(C) 36
(D) 24
(E) 18
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Re: permute with perfection

by x2suresh » Fri Feb 20, 2009 10:45 am
sanju09 wrote:A 4-letter code word consists of letters A, B, and C. If the code includes all the three letters, how many such codes are possible?

(A) 72
(B) 48
(C) 36
(D) 24
(E) 18
XABC --> X can be A/B/C

=3C1 * 4!/2! = 36

C
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by mals24 » Fri Feb 20, 2009 12:10 pm
C for me too

You could either repeat A (AABC), B (ABBC) or C (ABCC)

AABC can be arranged in 4!/2! ways

Since we have 3 such options total number of combination = 3*4!/2! = 36
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by pbanavara » Fri Feb 20, 2009 8:04 pm
Guys am terrible at permutations and combinations .. can someone please explain :

AABC can be arranged in 4!/2! ways - Is it because A is repeated twice ?

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by piyush_nitt » Fri Feb 20, 2009 8:05 pm
pbanavara wrote:Guys am terrible at permutations and combinations .. can someone please explain :

AABC can be arranged in 4!/2! ways - Is it because A is repeated twice ?

- pradeep
Yeah Pradeep ! You are right!
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IMO A

by chintudave » Fri Feb 20, 2009 10:44 pm
My Bad!

Each option was a duplicate.
Last edited by chintudave on Sat Feb 21, 2009 11:41 pm, edited 1 time in total.
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Re: IMO A

by sanju09 » Sat Feb 21, 2009 1:42 am
chintudave wrote:IMO A

The way i see it is that the question is asking about cases where A,B & C are used at least once and the forth letter could be repetition of A/B/C.

So i treat it as two groups.

Group 1: Total ways in which 3 letters can occupy 3 spaces is 6
Group 2: Total ways of selecting 1 alphabet from 3 alphabets for 1 space in the word is 3c1 i.e. 3

Total no of spaces available for Group 2 is 4c1 = 4

So total number of words are 6*3*4 = 72

Here are the actual options

AABC BABC CABC
AACB BACB CACB
ABAC BBAC CBAC
ABCA BBCA CBCA
ACAB BCAB CCAB
ACBA BCBA CCBA

AABC ABBC ACBC
AACB ABCB ACCB
BAAC BBAC BCAC
BACA BBCA BCCA
CAAB CBAB CCAB
CABA CBBA CCBA

ABAC ABBC ABCC
ACAB ACBB ACCB
BAAC BABC BACC
BCAA BCBA BCCA
CAAB CABB CACB
CBAA CBBA CBCA

ABCA ABCB ABCC
ACBA ACBB ACBC
BACA BACB BACC
BCAA BCAB BCAC
CABA CABB CABC
CBAA CBAB CBAC
B-) Are you sure all your actual options are unique, chintudave?
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by sumanr84 » Wed Jul 14, 2010 1:51 am
sanju09 wrote:A 4-letter code word consists of letters A, B, and C. If the code includes all the three letters, how many such codes are possible?

(A) 72
(B) 48
(C) 36
(D) 24
(E) 18
Ans: C

If A is the repeated letter, Total arrangment = 4!/2,
OR
if B is the repeated letter, Total arrangment = 4!/2
OR
if C is the repeated letter, Total arrangment = 4!/2

i.e. (4!/2) + (4!/2) + (4!/2 ) = 3*(4!/2 ) = 36 .
I am on a break !!
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by kvcpk » Wed Jul 14, 2010 2:35 am
It is a problem of placing (ABC) and (A or B or C) in 2 places.

First ABC can be arranged in 3! = 6 ways.

Now, fourth spot can be filled in 3 ways.
So 6*3 = 18

Now, the places of (ABC) and (A or B or C) can interchange.

So 2*18=36 ways.
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by GMATGuruNY » Wed Jul 14, 2010 3:44 am
A quick refresher for those less familiar with harder permutations.

How many ways can the letters ABC be arranged if no letters are repeated?

We have 3 choices for the 1st position.
We have 2 choices left for the 2nd position. (Because we used 1 of our 3 choices for the first position, leaving us only 2 choices for the 2nd position.)
We have 1 choice left for the last position. (Because we used 2 of our 3 choices for the first two positions, leaving us only 1 choice for the last position.)

Now we multiply the number of choices for each position:

3 * 2 * 1 = 3! = 6 possible arrangements.

So the total number of ways n elements can be arranged is n!

Now a harder question:

How many ways can the letters AAB be arranged if the letter A must appear exactly twice in the arrangement?

Normally, the number of arrangements would be 3!. But we have to account for the repeated A, which will reduce the number of unique arrangements. (If we wrote out all the possible unique arrangements, we'd see that there are only 3: AAB, ABA, and BAA.)

When an element is repeated, use the following formula:

(number of elements!)/(number of repetitions!)

In AAB we have 3 elements and 2 repetitions of the letter A, so we have to divide by 2!:

3!/2!= 3.

Another example:

How many ways can the letters ABBA be arranged?

We have 4 elements to arrange, but because we have 2 A's, we have divide by 2!, and because we have 2 B's, we have to divide again by 2!:

4!/(2! * 2!) = 6.

So moving onto the problem above:

A 4-letter code word consists of letters A, B, and C. If the code includes all the three letters, how many such codes are possible?

If all 3 letters have to be included in the code, one letter will need to be used twice. Our options are AABC, BBAC, and CCAB.

Number of ways to arrange AABC: 4!/2! = 12. (We divide by 2! because the A is repeated 2 times.)
Number of ways to arrange BBAC: 4!/2! = 12. (We divide by 2! because the B is repeated 2 times.)
Number of ways to arrange CCAB: 4!/2! = 12. (We divide by 2! because the C is repeated 2 times.)

Adding, we see that there are 12 + 12 + 12 = 36 possible arrangements.

The correct answer is C.
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by Nina1987 » Thu Feb 04, 2016 6:51 am
If the question didnt involve the restriction -"If the code includes all the three letters", would the answer be 3^4=81?
I am not able to VISUALIZE how that should be the answer. I arrived at that answer with a long winded process of considering each possible scenario such AAAA, AAAB, AABB, ABCA etc. But this approach will be too time consuming during the exam even if it were to give a correct answer.

Can some one help? thanks
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by MartyMurray » Thu Feb 04, 2016 7:30 am
Nina1987 wrote:If the question didnt involve the restriction -"If the code includes all the three letters", would the answer be 3^4=81?
I am not able to imagine how that should be the answer. Can some one help? thanks
Yes, if the code were just four letters, then you could determine the number of possibilities in the following way.

There are four slots, each of which can contain one of the three letters.

The first slot could contain A, B, or C. So there are three choices for the first slot.

For each of those three choices, there are three choices for the second slot.

First-->Second
A --> A, B or C
B --> A, B or C
C --> A, B or C

3 x 3 = 9 possible ways to fill the first two slots.

For each of those 9 setups, there are three ways to fill the third slot, and for each of those there area three ways to fill the fourth slot.

So there are 3 x 3 x 3 x 3 = 81 ways to fill all four slots.
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by Nina1987 » Thu Feb 04, 2016 7:39 am
Thanks for your reply Marty! it looks embarrassingly easy now. :)
I just tend to over think on certain problems I guess.
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by Jeff@TargetTestPrep » Thu Dec 14, 2017 9:46 am
sanju09 wrote:A 4-letter code word consists of letters A, B, and C. If the code includes all the three letters, how many such codes are possible?

(A) 72
(B) 48
(C) 36
(D) 24
(E) 18

We are given three letters, A, B, and C, and we must create a four-letter code in which all three letters are used. So, one letter must be repeated. Thus, we have the following three options:

1) A-B-C-A (if A is repeated)

2) A-B-C-B (if B is repeated)

3) A-B-C-C (if C is repeated)

Let's start with option 1:

We see that there are four total letters and two repeated As. Thus, that code can be selected in the following number of ways:

4!/2! = (4 x 3 x 2 x 1)/(2 x 1) = 4 x 3 = 12 ways

Since the second code, A-B-C-B, has two Bs rather than two As, we can create the second code in 12 ways. Likewise, since the third code, A-B-C-C, has two Cs rather than two As or two Bs, we can create the third code in 12 ways.

Thus, the code can be created in 12 + 12 + 12 = 36 ways.

Answer: C

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